Take any two positive numbers, say 4 and 16. Add them and halve it — you get their arithmetic mean: (4+16)/2=10. Multiply them and take the square root — you get their geometric mean: 4×16=8. Notice something? 10≥8. Try it with any other pair of positive numbers you like — the arithmetic mean is never smaller than the geometric mean. That simple, always-true observation is the Inequality of Means, usually written AM ≥ GM.
The precise statement
For two positive real numbers a and b:
AM=2a+b,GM=ab
2a+b≥ab
with equality if and only if a=b. If a=b, the inequality is strict.
Why it is always true
Start from a fact that can never fail: the square of any real number is non-negative.
(a−b)2≥0
Expand the left side:
a−2ab+b≥0
a+b≥2ab
Divide both sides by 2:
2a+b≥ab
That's the whole proof — no assumptions beyond a,b>0 (so that a,b are real numbers). Since (a−b)2=0 exactly when a=b, equality holds exactly when a=b.
Note
The inequality needs a,b≥0. For negative numbers, ab may not even be real, so the "GM" isn't defined there.
Worked example
Find the AM and GM of 9 and 25, and verify the inequality.
Step 1:AM=29+25=17
Step 2:GM=9×25=225=15
Step 3: Check: 17≥15✓ — and since 9=25, the inequality is strict, exactly as the rule predicts.
A useful consequence: inserting a mean between two numbers
If a and b are two positive numbers and G is inserted between them so that a,G,b form a Geometric Progression, then G=ab — precisely the geometric mean. Comparing this G against the arithmetic mean A=2a+b (the number that would sit between a and b in an Arithmetic Progression) is exactly an application of this inequality: A≥G always, so the AM-inserted term never sits below the GM-inserted term.
Watch out
A common slip is writing ab when a or b is negative, or applying the two-number formula directly to more than two numbers. For n positive numbers a1,a2,…,an, the generalised inequality is
Using the AM–GM inequality, the sum 4x+41−x is minimized when 4x=41−x, giving x=21 and a minimum value of 4.
Concept first.
When you see a sum of two positive terms where one is the reciprocal (or near-reciprocal) of the other, the AM–GM inequality is often the fastest route. Here 4x and 41−x are both positive for all real x, and their product is constant:
4x⋅41−x=4x+1−x=41=4.
That constant product is the key — it means the sum has a fixed lower bound.
Why AM–GM works here.
For any two non‑negative numbers a and b, the arithmetic mean is at least the geometric mean:
2a+b≥ab.
Equality holds exactly when a=b. So if we set a=4x and b=41−x, we get a direct bound on the sum.
Step‑by‑step solution
Apply AM–GM
Let a=4x and b=41−x. Then
24x+41−x≥4x⋅41−x.
Simplify the product
4x⋅41−x=4x+1−x=41=4.
So the right‑hand side becomes 4=2.
Obtain the inequality
24x+41−x≥2⇒4x+41−x≥4.
Find when equality occurs
AM–GM gives equality when a=b, i.e.
4x=41−x.
Since the base 4 is positive and not 1, we equate exponents:
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
Q.If both the roots of a quadratic equation x2+bx+c=0 are positive and b,c are non-zero real numbers, then 4bc is
(A) greater than or equal to b3
(B) greater than b3
(C) less than or equal to b3
(D) less than b3
›Reveal solutionSolution
With positive roots p,q: b=−(p+q), c=pq; the AM–GM inequality (p+q)2≥4pq rearranges to 4bc≥b3.
Set up with the roots. Let the two positive roots be p,q>0. By Vieta,
b=−(p+q)<0,c=pq>0.
Form the two expressions.
4bc=4(−(p+q))(pq)=−4pq(p+q),b3=−(p+q)3.
Compare. Divide the target inequality by −(p+q) (a negative number, so the inequality sign flips):
Q.If the harmonic mean of the roots of the equation 2x2−bx+(8−25)=0 is 4, then the value of b is
(A) 3
(B) 2
(C) 4−5
(D) 4+5
›Reveal solutionSolution
Use sum and product of roots from the quadratic's coefficients, plug into the harmonic-mean formula HM=sum2(product), and solve for b. The answer is 4−5.
Concept and Intuition
For two numbers α,β, the harmonic mean is HM=α+β2αβ=sum2(product). For a quadratic px2+qx+r=0 with roots α,β: sum =−q/p and product =r/p (Vieta's formulas) — so the harmonic mean of the roots can be written directly in terms of the coefficients without ever solving for the roots themselves.
Step-by-Step Solution
For 2x2−bx+(8−25)=0: comparing to px2+qx+r=0 gives p=2, q=−b, r=8−25.
Sum of roots =α+β=−q/p=2b.
Product of roots =αβ=r/p=28−25.
Harmonic mean =α+β2αβ=2b2⋅28−25=b2(8−25) — the 2 in numerator and denominator cancels cleanly. …
Q.If a, b, c are distinct positive real numbers and a2+b2+c2=1, the value of ab+bc+ca is
(A) less than 1
(B) greater than 1
(C) equals to 1
(D) any real number
›Reveal solutionSolution
This tests the classical inequality a2+b2+c2≥ab+bc+ca (with strict inequality when the numbers are distinct), applied to the constraint a2+b2+c2=1.
Concept and Intuition
The key inequality here comes from the fact that a sum of squares is always non-negative, and it is strictly positive when the numbers involved are not all equal. Since a,b,c are explicitly stated to be distinct, we get a strict inequality, which pins down ab+bc+ca to be strictly less than a2+b2+c2.
Step-by-Step Solution
Consider the identity (a−b)2+(b−c)2+(c−a)2=2(a2+b2+c2)−2(ab+bc+ca).
Since a,b,c are distinct, at least one of (a−b),(b−c),(c−a) is nonzero, so the left-hand side (a sum of squares) is strictly positive: (a−b)2+(b−c)2+(c−a)2>0.