Q.For any two positive real numbers $a$ and $b$, prove that $\dfrac{a}{b}+\dfrac{b}{a} \geq 2$.

The Intuition: Why the Average Isn't Always the Middle

Imagine you have two numbers: 2 and 8. Their ordinary arithmetic mean (AM) is $(2+8)/2 = 5$. That feels natural — the number halfway between them on a number line.

Now think about the geometric mean (GM). For 2 and 8, it's $\sqrt{2 \times 8} = \sqrt{16} = 4$. That's smaller than 5. Why?

The geometric mean cares about multiplicative spacing, not additive spacing. On a number line, 2 and 8 are 6 units apart, so the arithmetic mean sits right in the middle. But if you think in terms of ratios: 2 multiplied by 4 gives 8. The geometric mean is the number that, when multiplied by itself, gives the same product — it's the "multiplicative middle." And that middle is always pulled toward the smaller number when the numbers are unequal.

The Precise Statement

For any two positive real numbers $a$ and $b$:

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

Where:

This inequality is not just a curiosity — it's a fundamental result that holds for any set of positive numbers, not just two. For $n$ positive numbers $x_1, x_2, \dots, x_n$:

Why It's True: A Simple Proof

Start with any two positive numbers $a$ and $b$. Consider $(\sqrt{a} - \sqrt{b})^2$.

Since a square is always non-negative:

Expand:

Rearrange:

Divide by 2:

That's it. The equality case is when $\sqrt{a} - \sqrt{b} = 0$, i.e., $a = b$.

A Concrete Example

Take $a = 4$, $b = 9$.

• AM = $(4+9)/2 = 6.5$
• GM = $\sqrt{4 \times 9} = \sqrt{36} = 6$

AM (6.5) > GM (6). The gap is small because the numbers are close.

Now take $a = 1$, $b = 100$.

• AM = $(1+100)/2 = 50.5$
• GM = $\sqrt{1 \times 100} = 10$

The gap is huge — the geometric mean is far more sensitive to the smaller number.

Why This Matters

The AM–GM inequality is a workhorse in optimization. If you have a fixed sum, the product is maximized when the numbers are equal. If you have a fixed product, the sum is minimized when the numbers are equal. This lets you solve problems like "Find the minimum value of $x + \frac{1}{x}$ for $x > 0$" without calculus — just apply AM–GM to $x$ and $1/x$.

The relationship between AM and GM is not just a fact to memorize — it's a lens through which you see that averaging additively and averaging multiplicatively give different answers, and the additive average is always at least as large.