Q.If A is the arithmetic mean and $G_1, G_2$ be two geometric means between any two numbers, then prove that $2A = \dfrac{G_1^2}{G_2} + \dfrac{G_2^2}{G_1}$.

Inequality of Means

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: $\sqrt{4 \times 16} = 8$. Notice something? $10 \ge 8$. 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 $b$:

with equality if and only if $a = b$. If $a \ne b$, 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.

Expand the left side:

Divide both sides by $2$:

That's the whole proof — no assumptions beyond $a, b > 0$ (so that $\sqrt{a}, \sqrt{b}$ are real numbers). Since $(\sqrt a - \sqrt b)^2 = 0$ exactly when $\sqrt a = \sqrt b$, equality holds exactly when $a = b$.

Worked example

Find the AM and GM of $9$ and $25$, and verify the inequality.

Step 1: $\text{AM} = \dfrac{9+25}{2} = 17$

Step 2: $\text{GM} = \sqrt{9 \times 25} = \sqrt{225} = 15$

Step 3: Check: $17 \ge 15$ ✓ — and since $9 \ne 25$, 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 $G = \sqrt{ab}$ — precisely the geometric mean. Comparing this $G$ against the arithmetic mean $A = \dfrac{a+b}{2}$ (the number that would sit between $a$ and $b$ in an Arithmetic Progression) is exactly an application of this inequality: $A \ge G$ always, so the AM-inserted term never sits below the GM-inserted term.