Did you know? A tangent line can prove an inequality for infinitely many numbers. Draw the graph of y=eˣ and its tangent at x=0. The tangent is y=1+x, and the exponential curve never falls below it. That single geometric fact gives eˣ≥1+x for every real x, with equality at zero.
For JC2 students and families investigating a JC2 H3 Mathematics tutor in Bukit Timah, this is a gateway to convexity and tangent-line inequality proofs. The 2027 Singapore–Cambridge H3 Mathematics (9820) syllabus explicitly includes AM–GM, Cauchy–Schwarz, the triangle inequality and H2 calculus as assumed knowledge. Convexity and Jensen-type arguments are connected enrichment methods, not separately named additional H3 topics.
Quick answer: For a differentiable convex function f on an interval, each tangent line is a global lower bound: f(x)≥f(a)+f′(a)(x−a) for points x,a in that interval. A differentiable concave function has the opposite inequality. Checking the domain, curvature and equality conditions turns this idea into a rigorous proof instead of a guess from a graph.
Quick navigation
- Convex and concave in plain language
- Why a tangent line bounds the whole curve
- Proof of the tangent inequality
- Exponential inequality e^x ≥ 1+x
- Logarithmic inequality ln x ≤ x−1
- Square roots and AM–GM
- Jensen’s midpoint and weighted inequalities
- Optimisation and equality conditions
- How to check the sign and domain
- Original worked inequality clinics
- Common mathematical errors
- FAQs and syllabus references
What convex and concave really mean
A function is convex on an interval when the straight segment joining any two points on its graph lies on or above the graph between those points. Equivalently, for 0≤t≤1, f(tx+(1−t)y)≤tf(x)+(1−t)f(y) for any x,y in the interval.
Concavity reverses this inequality: the graph sits on or above its chords. This describes curvature through a mathematical condition rather than relying on how a curve happens to look on a computer screen.
Examples of convex functions on the real line include f(x)=x², f(x)=x⁴ and f(x)=eˣ. A useful strictly concave function on the positive real line is f(x)=ln x. The square-root function is concave on its nonnegative domain.
For a twice-differentiable function on an interval, f″(x)≥0 throughout the interval is a sufficient condition for convexity. If f″(x)≤0 throughout, it is a sufficient condition for concavity. The inequalities refer to the entire relevant interval, not one sample point.
Be careful about converses. A convex function need not have a second derivative everywhere, so convexity does not imply the existence of a globally nonnegative second derivative. For instance f(x)=|x| is convex but not differentiable at zero.
The geometric terms are useful only when linked to exact inequalities. If the problem asks for a proof, state the interval, derivative or chord property and logical reason why the relevant bound holds.
Why a convex tangent gives a global lower bound
Suppose f is differentiable and convex on an interval, and a belongs to its interior. The tangent line at x=a has equation y=f(a)+f′(a)(x−a). The convexity property ensures the graph of f stays on or above this line throughout the interval.
This is different from a local linear approximation, which merely estimates a function near a point. For convex functions, the tangent is not just close near a; it is a guaranteed global lower bound on the entire interval.
One way to see it is through increasing slopes. For x>a, a convex function’s secant slope [f(x)−f(a)]/(x−a) is at least f′(a). Multiply by x−a>0 to obtain f(x)≥f(a)+f′(a)(x−a).
For x<a, the secant slope is at most f′(a), but multiplying by x−a<0 reverses the inequality, giving the same lower-bound conclusion. Both sides of a have been covered.
For differentiable concave f, the tangent lies above the graph: f(x)≤f(a)+f′(a)(x−a). This is especially useful for logarithms and square roots, where an upper bound may be easier to work with than the original nonlinear expression.
Equality always holds at the tangent point x=a. For a strictly convex function it holds nowhere else when the tangent is taken at an interior point. That equality condition is often the step needed to turn a bound into an actual minimum.
A well-written H3 proof does not simply say ‘by convexity’ if the function’s convexity or the relevant interval has not been established. Identify the function and the condition that makes the tangent theorem available.
Prove the tangent inequality through derivatives
Let f be twice differentiable on an interval with f″(x)≥0. Fix a point a. Define g(x)=f(x)−[f(a)+f′(a)(x−a)], the gap between the curve and its tangent at a.
The gap satisfies g(a)=0 and g′(a)=0. Its second derivative is g″(x)=f″(x)≥0. Therefore g′ is nondecreasing, which means g′(x)≤0 to the left of a and g′(x)≥0 to the right of a.
Hence g decreases towards a from the left and increases away from a on the right. It attains a global minimum at a, where its value is zero. Thus g(x)≥0 for every x in the interval and the tangent inequality follows.
This argument provides a proof with a clear direction: derivative condition, monotonicity, minimum of the gap, global inequality. It also explains why checking f″(a)≥0 at just one point is insufficient; the sign must be controlled on the relevant interval.
If f″(x)≤0 throughout, the same argument gives g″≤0 and the tangent becomes a global upper bound. In both settings, differentiability, the domain and any endpoints matter.
Students with strong H2 calculus can often build such an argument directly without requiring an abstract theorem about convex functions. That is a practical route from familiar derivatives to H3 inequality reasoning.
Full worked proof: eˣ≥1+x
Consider f(x)=eˣ for real x. Its first and second derivatives are both eˣ, which is strictly positive. Therefore f is strictly convex on the whole real line.
At a=0, we have f(0)=1 and f′(0)=1. The tangent is y=1+x. By the tangent inequality, eˣ≥1+x for all real x, with equality exactly at x=0.
To prove the same result using a calculus function, let g(x)=eˣ−x−1. Then g′(x)=eˣ−1. It is negative for x<0, zero at x=0 and positive for x>0. Thus g has global minimum zero at x=0.
These are two valid proof routes: one cites convexity after establishing its derivative criterion, while the other directly shows the gap is nonnegative. The direct proof may be shorter for students who have not formally studied convexity.
Changing x to −t gives e^(−t)≥1−t for every real t. For t≥0 this provides a lower estimate for exponential decay, although it can be weak or even negative for large t.
The original statement is global, not merely a small-x approximation. For x=−2, e^(−2)≈0.135 remains above 1−2=−1. A negative tangent estimate is allowed because it is a lower bound, not an attempted approximation equal to the function.
The equality condition matters in optimisation. The inequality tells us eˣ−x≥1 for all real x, with equality at zero. Therefore eˣ−x has minimum value one, attained at x=0.
Full worked proof: ln x≤x−1
Let f(x)=ln x on the positive real line. Its second derivative is f″(x)=−1/x², strictly negative for every x>0. Therefore f is strictly concave on that interval.
At a=1, the value is ln1=0 and derivative 1/x=1. The tangent line is y=x−1. Concavity gives ln x≤x−1 for every x>0, with equality only at x=1.
An alternative calculus proof studies h(x)=x−1−ln x. Its derivative is h′(x)=1−1/x=(x−1)/x. The derivative is negative on (0,1), zero at one and positive on (1,∞). Thus h has minimum zero at x=1.
The domain x>0 cannot be omitted: ln x is not a real number at zero or for negative x. A well-presented solution states this condition before taking derivatives or substituting into the inequality.
This inequality is a useful foundation for comparing logarithmic and polynomial growth. For example, apply ln t≤t−1 to t=√x with x≥1. Then ln x=2ln√x≤2(√x−1). Divide by x to get 0≤(ln x)/x≤2/√x, showing the ratio tends to zero.
The same tangent result can be restated as x−ln x≥1 for positive x, equality at one. It gives a concrete global minimum without needing to solve a difficult optimisation problem.
Notice how the concavity argument is the mirror image of the exponential argument. Convex curves lie above their tangents; concave curves lie below them.
Square roots and the arithmetic–geometric mean
Consider f(x)=√x for x>0. Its second derivative is negative, so it is concave on the positive domain. At x=1, f(1)=1 and f′(1)=1/2, giving tangent y=(x+1)/2.
Thus √x≤(x+1)/2 for x>0, with equality at x=1. The inequality also holds at x=0 by continuity. Rearranging gives the familiar AM–GM relation between x and one.
To obtain AM–GM for any a,b>0, set x=a/b. The inequality √(a/b)≤(a/b+1)/2 becomes 2√(ab)≤a+b after multiplying by b. Equality occurs exactly when a=b.
This is a valuable connection: a standard inequality listed in the H3 syllabus can be derived from a tangent to a concave function. The geometric viewpoint helps explain not only the bound but also its equality case.
More generally, concavity of ln x gives ln[ta+(1−t)b]≥t ln a+(1−t)ln b for a,b>0 and 0≤t≤1. Exponentiating produces weighted AM–GM: ta+(1−t)b≥aᵗb^(1−t).
At t=1/2 we recover ordinary AM–GM. For strict positive unequal a,b and 0<t<1, the inequality is strict because log is strictly concave.
These extensions should be labelled enrichment when discussing syllabus alignment. The syllabus explicitly names AM–GM; it does not independently list every theorem concerning convexity or logarithmic Jensen inequalities.
Jensen’s inequality and midpoint reasoning
Convexity itself gives a two-point weighted inequality: f(tx+(1−t)y)≤tf(x)+(1−t)f(y). This is sometimes called the two-point form of Jensen’s inequality. The weights must be nonnegative and sum to one.
Set t=1/2 and f(z)=z². Then [(x+y)/2]²≤(x²+y²)/2. Multiply by four and rearrange to obtain x²+y²≥2xy, equivalent to (x−y)²≥0.
Thus a familiar algebraic square inequality can also be understood as the convexity of z². It is useful to compare the two proofs and ask which makes the equality condition most transparent.
For the convex exponential, Jensen gives e^((a+b)/2)≤(eᵃ+eᵇ)/2 for all real a,b. This is AM–GM applied to the positive numbers eᵃ and eᵇ. Equality holds at a=b.
For positive x,y, the convex function f(z)=1/z gives 1/[(x+y)/2]≤[1/x+1/y]/2. Rearranging yields 1/x+1/y≥4/(x+y), a useful inequality relating arithmetic and harmonic means.
Jensen’s inequality extends to more points and weights, but a complete proof must state the convexity domain and the weights. A negative weight would no longer be covered by the usual convex-combination condition.
Students should not use a theorem name as a substitute for defining the function and its curvature. A well-argued choice of f and weights is the mathematical work in this method.
Optimisation, equality and constraints
A convex differentiable function has a global minimum at any interior point where its derivative is zero, provided the point lies in the stated interval. The tangent at such a stationary point is horizontal and forms a global lower bound.
Consider f(x)=x²−4x+7 on the real line. Its second derivative is two, positive everywhere. Solve f′(x)=2x−4=0 to get x=2. The minimum is f(2)=3.
Completing the square gives the same proof: f(x)=(x−2)²+3≥3, equality at x=2. This direct algebraic route is shorter, while the convexity route extends to more general differentiable functions.
Now restrict the domain to x≥3. The unconstrained minimiser x=2 is no longer allowed. Because f is increasing for x>2, the constrained minimum occurs at the boundary x=3 with value four. The correct optimum depends on the feasible set.
Do not claim every strictly convex function has a global minimum on the real line. The exponential f(x)=eˣ is strictly convex but has no attained minimum on ℝ: it approaches zero as x→−∞ without ever reaching zero.
This distinction separates strict convexity, uniqueness and existence. Strict convexity gives at most one minimiser under suitable conditions; it does not by itself guarantee that a minimiser exists on an arbitrary domain.
For practical problems, the equality condition should always be checked against the constraints. A bound that cannot be reached may still be a valid inequality, but it is not an attained optimum.
Domain and sign: the most frequent proof mistakes
The tangent theorem requires convexity or concavity on the entire interval being used. The function f(x)=x³ has f″(x)=6x, changing sign at zero. Its tangent at zero is y=0, but x³ lies below that line for negative x, so the claimed global convex lower bound would be false.
The reciprocal function f(x)=1/x is convex on x>0, where f″(x)=2/x³>0. It is not convex over a domain crossing zero because it is undefined at zero and the second derivative changes sign on negative inputs.
At x=1 the tangent to 1/x is 2−x, so 1/x≥2−x for x>0. This is an exact inequality despite the right-hand side becoming negative for x>2; a lower bound is allowed to be weak.
Applying ln x≤x−1 with x≤0 is invalid because ln x is not real there. Applying √x≤(x+1)/2 to negative real x is likewise undefined. Before deriving an inequality, verify that all function arguments belong to the correct domain.
Multiplying an inequality by an expression of unknown sign can reverse its direction. The weighted Jensen inequality avoids this by requiring nonnegative weights, while AM–GM requires nonnegative inputs. Stating these conditions is as important as manipulating the formula correctly.
Another danger is mistaking a local stationary point for a global maximum or minimum. For a convex function, a stationary point gives a global minimum, not a global maximum. For a nonconvex function, the derivative alone is insufficient.
A strong H3 solution explains the choice of theorem, the interval and the equality case. The AM–GM and equality guide develops the prescribed inequality itself; this article’s distinct focus is tangent-line and curvature-based methods.
Two extended investigations: the right function changes the whole proof
Investigation one — the best bound for a sum of reciprocals. Let a,b be positive real numbers. We want a lower bound on 1/a+1/b in terms of their sum a+b. A direct common denominator gives (a+b)/(ab), but the best universal relationship is not immediately obvious.
Choose f(x)=1/x, which is strictly convex for x>0 because f″(x)=2/x³>0. By the midpoint convexity inequality, f((a+b)/2)≤[f(a)+f(b)]/2. Substituting gives 2/(a+b)≤(1/a+1/b)/2, so 1/a+1/b≥4/(a+b).
Equality holds when a=b, since 1/x is strictly convex on positive reals. Check with a=b=2: both sides are one. With a=1,b=3, the left side is 4/3 and the right is one, so the inequality is strict.
Now change the condition from positive reals to arbitrary nonzero reals. The same argument no longer follows because the interval between a and b might cross zero, where f is undefined. The positivity condition is part of the theorem, not an incidental detail.
There is also an elementary algebraic proof: subtract 4/(a+b) from (a+b)/(ab). The result is (a−b)²/[ab(a+b)], nonnegative when a,b>0. The two proofs are complementary. The algebra shows the exact nonnegative gap; the convexity proof explains the broader pattern.
Investigation two — a minimum without a complicated graph. Consider g(x)=x−ln x for x>0. Its second derivative is 1/x²>0, so it is strictly convex. Its first derivative is 1−1/x, vanishing only at x=1. Thus x=1 is its unique global minimiser, with g(1)=1.
Alternatively, tangent concavity of ln at one gives ln x≤x−1, rearranging directly to x−ln x≥1. This avoids calculating a second derivative after the standard bound is known, and equality occurs at one.
Suppose the domain changes to x≥2. The old equality case x=1 is no longer permitted. Since g′(x)>0 for x>1, g increases throughout the new domain; the minimum occurs at its boundary x=2 and equals 2−ln2. A global inequality can remain true while no longer give the attained constrained minimum.
These two investigations show why H3 Mathematics rewards proof selection. A tangent, a square difference, a derivative sign or an inequality theorem might all address the same problem. The strongest solution is the one whose assumptions and equality case are easiest to justify.
Original convexity and inequality clinics
1. Tangent to x² at x=1
Worked result: The tangent is y=2x−1 and x²≥2x−1, since their difference is (x−1)².
Proof check: An exact square proves a convex tangent lower bound without relying on a diagram.
2. Tangent to x² at x=2
Worked result: The tangent is y=4x−4, so x²≥4x−4 for every real x.
Proof check: Equality occurs at x=2 and the difference is (x−2)².
3. Tangent to x² at x=a
Worked result: For any real a, x²≥2ax−a² follows from (x−a)²≥0.
Proof check: A general parameter reveals the entire family of quadratic tangent inequalities.
4. Convexity of x⁴
Worked result: The second derivative 12x² is nonnegative for every real x, so x⁴ is convex.
Proof check: A derivative test must hold across the entire interval, not at one point.
5. Concavity of −x²
Worked result: The second derivative is −2, so −x² is concave and stays below its tangents.
Proof check: Reversing curvature reverses the tangent inequality.
6. A concave tangent at one
Worked result: The tangent to −x² at x=1 is y=1−2x, so −x²≤1−2x.
Proof check: The difference reduces to the nonnegative square (x−1)².
7. Exponential tangent at zero
Worked result: eˣ≥1+x for every real x; equality is x=0.
Proof check: Strict convexity supplies a global lower bound rather than just a local approximation.
8. Exponential tangent at one
Worked result: eˣ≥ex for every real x, with equality x=1.
Proof check: The tangent slope and intercept at x=1 are both determined by e.
9. Exponential tangent at minus one
Worked result: eˣ≥(x+2)/e for all real x, with equality x=−1.
Proof check: A tangent point can be chosen strategically to match the desired linear expression.
10. Minimum of eˣ−x
Worked result: The minimum is one at x=0 because eˣ−x≥1.
Proof check: A sharp bound must include an allowed equality case.
11. Negative input in exponential inequality
Worked result: At x=−2, e^(−2)>0 while 1+x=−1, so the bound holds.
Proof check: A lower bound can be negative and still be mathematically correct.
12. Logarithmic tangent at one
Worked result: ln x≤x−1 for every x>0, with equality at one.
Proof check: Concavity produces an upper bound, opposite to the convex case.
13. Logarithmic tangent at two
Worked result: ln x≤ln2+(x−2)/2 for x>0.
Proof check: The tangent formula uses f(2)=ln2 and f′(2)=1/2.
14. A transformed logarithm inequality
Worked result: ln x≤2(√x−1) for every positive x.
Proof check: Apply ln t≤t−1 to t=√x, then multiply by two.
15. Logarithm at one-half
Worked result: ln(1/2)≤−1/2, a true strict inequality.
Proof check: A numerical test can catch a sign mistake but does not replace the global proof.
16. Reciprocal tangent at one
Worked result: For x>0, 1/x≥2−x, equality x=1.
Proof check: Strict convexity of 1/x holds only on a positive interval.
17. Reciprocal tangent at two
Worked result: For x>0, 1/x≥1−x/4, equality x=2.
Proof check: Substitute f(2)=1/2 and f′(2)=−1/4 into the tangent formula.
18. Square-root tangent at one
Worked result: For x≥0, √x≤(x+1)/2, with equality x=1.
Proof check: The inequality extends to zero by continuity from the positive domain.
19. Square-root tangent at four
Worked result: For x≥0, √x≤1+x/4, equality x=4.
Proof check: A different tangent point produces another global upper bound.
20. Ordinary AM–GM
Worked result: For a,b≥0, (a+b)/2≥√(ab), equality when a=b.
Proof check: Apply square-root concavity or use (√a−√b)²≥0.
21. Weighted AM–GM
Worked result: For a,b>0 and 0≤t≤1, ta+(1−t)b≥aᵗb^(1−t).
Proof check: Concavity of ln turns a weighted average into a geometric bound.
22. Weights at endpoints
Worked result: When t=0 or t=1, weighted AM–GM is equality regardless of a and b.
Proof check: Strict equality conditions require the weight to be genuinely between zero and one.
23. A negative weight failure
Worked result: For f(x)=x², x=1,y=3,t=2 gives f(−1)=1 but 2f(1)−f(3)=−7.
Proof check: Jensen’s convex-combination inequality cannot be used with arbitrary negative weights.
24. Quadratic midpoint inequality
Worked result: For real a,b, [(a+b)/2]²≤(a²+b²)/2.
Proof check: The gap is (a−b)²/4, nonnegative for every real pair.
25. A numerical midpoint check
Worked result: For f(x)=x² and points one and three, f(2)=4≤(1+9)/2=5.
Proof check: A check illustrates but does not prove the universal convexity inequality.
26. Exponential midpoint
Worked result: For all real a,b, e^((a+b)/2)≤(eᵃ+eᵇ)/2.
Proof check: This is AM–GM applied to two positive exponential values.
27. Reciprocal midpoint
Worked result: For positive a,b, 1/a+1/b≥4/(a+b).
Proof check: The inequality expresses the convexity of 1/x on positive reals.
28. An exact reciprocal gap
Worked result: The difference 1/a+1/b−4/(a+b) equals (a−b)²/[ab(a+b)] for a,b>0.
Proof check: An explicit nonnegative quotient verifies both the inequality and equality case.
29. Quadratic optimisation
Worked result: x²−4x+7=(x−2)²+3 has minimum three at x=2.
Proof check: Completing the square gives a global bound with a visible equality case.
30. Constrained quadratic optimisation
Worked result: For x≥3, x²−4x+7 is minimised at x=3 with value four.
Proof check: A minimiser outside the feasible set cannot answer the constrained problem.
31. A reciprocal-plus-linear objective
Worked result: For x>0, x+4/x≥4, with equality at x=2.
Proof check: The two positive terms have product four, so AM–GM is sharp.
32. Minimising x−ln x
Worked result: For x>0, x−ln x≥1, with equality at x=1.
Proof check: Either logarithmic concavity or a derivative sign proves the exact global minimum.
33. Convex but no minimiser
Worked result: The function eˣ is strictly convex on ℝ but does not attain its infimum zero.
Proof check: Strict convexity does not by itself guarantee existence of a global minimum.
34. Stationary convex function
Worked result: If a differentiable convex function has f′(a)=0 at an interior point, a is a global minimiser.
Proof check: Its tangent at a is horizontal and bounds every other function value below.
35. A boundary minimum
Worked result: The convex function x² on [1,∞) is minimised at one even though f′(1)=2.
Proof check: A constrained minimiser can occur at a boundary without zero derivative.
36. Cubic tangent failure
Worked result: For f(x)=x³, the tangent at zero is y=0, but the curve lies below it for negative x.
Proof check: f″(x)=6x changes sign, so no global convex tangent inequality applies on ℝ.
37. Concavity of sine on a range
Worked result: On [0,π], f(x)=sin x is concave because f″(x)=−sin x≤0.
Proof check: The sign of curvature is interval-dependent.
38. Sine tangent at zero
Worked result: For 0≤x≤π, sin x≤x by concavity and the tangent at zero.
Proof check: The bound is tied to a domain where the concavity condition has been established.
39. Logarithmic domain
Worked result: The expression ln x is real only for x>0.
Proof check: An elegant inequality with a logarithm cannot be extended to negative inputs without a new definition.
40. Reciprocal domain
Worked result: The function 1/x is convex on x>0, where f″=2/x³>0.
Proof check: Its behaviour cannot be treated as one convex function across the singularity at zero.
41. Checking equality
Worked result: A candidate bound is an attained optimum only if an allowed input makes it an equality.
Proof check: A correct bound alone does not prove the maximum or minimum value is reached.
A seven-session JC2 practice plan
Session one connects familiar quadratic inequalities to the idea of convexity. Have the student derive x²≥2ax−a² by completing a square, then identify the tangent point a and its equality condition.
Session two introduces the derivative test for convexity and concavity. Compare x², x⁴, eˣ, ln x and √x, including their correct domains. Ask why a second derivative with changing sign makes a global claim unsafe.
Session three derives the tangent-line inequality from monotone slopes. The learner should explain the cases x>a and x<a and the reversal of an inequality when multiplied by a negative quantity.
Session four proves eˣ≥1+x and ln x≤x−1 in two independent ways: with the tangent theorem and with a direct derivative sign. Discuss which proof more clearly exposes the equality case.
Session five develops AM–GM through square-root concavity and derives weighted AM–GM from logarithmic concavity, explicitly stating positive inputs and nonnegative weights.
Session six applies convexity to optimisation. Contrast the unconstrained minimum of x²−4x+7 with the same function on x≥3, and compare the reciprocal inequality with its algebraic square-gap proof.
Session seven presents an unfamiliar inequality. The student must select a function, identify its curvature on the permitted domain, choose an effective tangent point and establish equality or sharpness without relying only on a graph.
Parents can ask which curve is being used, whether the tangent sits above or below it, and where equality occurs. A clear explanation of those three ideas often reveals deeper learning than repeated calculation.
The immutable eduKateSG Mathematics reference near Sixth Avenue MRT describes a three-student Secondary 1 model, not confirmation of JC2 H3 classes. Enquire directly about current advanced Mathematics tutoring.
Frequently asked questions
Is convexity separately named in the 2027 H3 9820 outline?
The syllabus names AM–GM, Cauchy–Schwarz, triangle inequalities and H2 calculus. Convexity and Jensen methods are connected enrichment applications, not separately listed additional H3 content.
Why does a tangent give a global bound?
For a differentiable convex function on an interval, monotone slopes place its graph on or above every tangent. Concavity reverses the bound.
What is the easiest convexity check?
When a function is twice differentiable on an interval, showing f″≥0 throughout establishes convexity. Check the interval and avoid assuming all convex functions have second derivatives.
Is e^x≥1+x valid for negative x?
Yes, for every real x, with equality at zero. The exponential curve is strictly convex and the tangent at zero is y=1+x.
Why is ln x≤x−1 restricted to positive x?
The real logarithm and its derivatives are defined only for x>0. Its concavity supplies the tangent upper bound at x=1.
How does convexity connect to AM–GM?
The square-root function is concave and ln is strictly concave on positive inputs. Their chord or weighted tangent inequalities yield ordinary and weighted AM–GM.
Does a strictly convex function always attain a minimum?
No. Strict convexity controls the shape and possible uniqueness but does not guarantee existence. For example eˣ has infimum zero on ℝ and never reaches it.
How are equality conditions checked?
Identify the tangent point or the equality condition of the inequality, then verify it belongs to the feasible domain. A bound without an attained equality case may not be the requested optimum.
Why not apply Jensen with negative weights?
The standard convexity statement requires nonnegative weights summing to one. Negative weights can reverse or invalidate the inequality.
How should an H3 tutor assess a proof?
Ask for the selected function, the interval, the derivative or convexity justification, the tangent or chord equation and a precise equality condition.
Official source and companion mathematical reading
The 2027 Singapore–Cambridge H3 Mathematics 9820 syllabus names the principal additional inequalities, assumes H2 calculus and emphasises mathematical proof and non-routine reasoning. Convexity here is a connected enrichment method, not a separate prescribed chapter.
For adjacent techniques, read AM–GM inequality and equality, Cauchy–Schwarz and triangle inequalities and modelling constraints. Together they build the habit of choosing a valid bound and proving when it is sharp.
A final two-minute inequality audit
When a problem looks difficult, begin by asking what kind of inequality it seeks: a global lower bound, a global upper bound, a local approximation or an optimum under constraints. Decide whether the selected function is convex or concave on the entire permissible interval. Write the tangent at the intended equality point and check whether its direction agrees with the curvature. Then verify every step of the supporting derivative argument rather than relying on a plot that happens to look convincing.
Before giving the final answer, inspect the equality case. A sharp inequality is most useful when the equality value lies within the problem’s domain. If that point is excluded, the same inequality may remain true while the constrained optimum moves to a boundary or fails to be attained. This is why domain, curvature, tangent point and equality are four independent pieces of an excellent proof. Together they turn an attractive geometric idea into precise mathematical communication.