Q.Prove that the arithmetic mean of two positive real numbers is never less than their geometric mean, and verify your proof numerically for $a=9$, $b=16$.

[!TLDR] $(\sqrt a-\sqrt b)^2\ge0$ expands to $a+b\ge2\sqrt{ab}$, i.e. $A\ge G$; for $a=9,b=16$: $A=12.5\ge G=12$. [!ANSWER] $A\ge G$ always, proved via $(\sqrt a-\sq …

For positive reals $a,b$: since the square of any real number is non-negative, $(\sqrt a-\sqrt b)^2\ge0$. Expanding: $a-2\sqrt{ab}+b\ge0 \Rightarrow a+b\ge2\sqrt{ab} \Rightarrow \dfrac{a+b}{2}\ge\sqrt{ab}$, i.e. $A\ge G$, with equality iff $\sqrt a=\sqrt b$ iff $a=b$. Numerical check with $a=9,b=16$: $A=\dfrac{9+16}{2}=12.5$; $G=\sqrt{9\times16}=\sqrt{144}=12$. Indeed $12. …

Square the real number $\sqrt a-\sqrt b$ (always $\ge0$), expand, and rearrange to isolate $\frac{a+b}{2}\ge\sqrt{ …

Squaring $\sqrt a - \sqrt b$ incorrectly as $a-b$ (instead of $a-2\sqrt{ab}+b$) is a common a …