Prove AM ≥ GM for two positive numbers

Question

Prove that for any two positive real numbers $a$ and $b$ , the arithmetic mean is greater than or equal to the geometric mean:

\frac{a+b}{2} \geq \sqrt{ab}

with equality if and only if $a = b$ .

Solution — Step by Step

For any real number, a square is always $\geq 0$ . So:

(\sqrt{a} - \sqrt{b})^2 \geq 0

This holds for all real $\sqrt{a}$ and $\sqrt{b}$ (which are real since $a, b > 0$ ).

(\sqrt{a})^2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b})^2 \geq 0

a - 2\sqrt{ab} + b \geq 0

a + b \geq 2\sqrt{ab}

\frac{a+b}{2} \geq \sqrt{ab}

This is exactly AM $\geq$ GM. ✓

Equality holds when $(\sqrt{a} - \sqrt{b})^2 = 0$ , which means $\sqrt{a} = \sqrt{b}$ , i.e., $a = b$ .

So AM = GM if and only if $a = b$ .

Why This Works

The proof reduces a non-obvious inequality to the obvious fact that squares are non-negative. The substitution $\sqrt{a}$ and $\sqrt{b}$ (rather than $a$ and $b$ ) is the key insight — it makes the algebra work cleanly.

This inequality is fundamental in mathematics because it tells us that the AM is a pessimistic upper bound on the “average” in the sense of multiplication. The geometric mean is always pulled down by any difference between $a$ and $b$ .

Intuition: If you have a rectangle with sides $a$ and $b$ and perimeter $2(a+b)$ , the area $ab$ is maximised when the rectangle is a square ( $a = b$ ). This is exactly AM-GM at work.

Alternative Method — Algebraic Manipulation

Start directly from what we want to prove and show it’s equivalent to something obviously true.

We want: $\dfrac{a+b}{2} \geq \sqrt{ab}$

Squaring both sides (valid since both sides are positive):

\left(\frac{a+b}{2}\right)^2 \geq ab

\frac{(a+b)^2}{4} \geq ab

(a+b)^2 \geq 4ab

a^2 + 2ab + b^2 \geq 4ab

a^2 - 2ab + b^2 \geq 0

(a-b)^2 \geq 0

This is always true. Working backwards, every step is reversible, so AM $\geq$ GM is proved.

Both proofs are valid for JEE and CBSE. The first method (starting from $(\sqrt{a}-\sqrt{b})^2 \geq 0$ ) is cleaner for direct proof. The second method works backwards — fine for exams, but in a formal proof, ensure all your steps are reversible or rewrite it forwards.

Common Mistake

Some students try to prove AM $\geq$ GM by assuming the result and squaring: “Let $\frac{a+b}{2} \geq \sqrt{ab}$ . Squaring, $\frac{(a+b)^2}{4} \geq ab \ldots$ and so $(a-b)^2 \geq 0$ which is true.” This looks like circular reasoning unless you explicitly state that all steps are “if and only if” (reversible). Always begin with something known to be true ( $(\sqrt{a}-\sqrt{b})^2 \geq 0$ ) and work forward to derive the inequality.

Question

Solution — Step by Step

Why This Works

Alternative Method — Algebraic Manipulation

Common Mistake

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