Let $u$ and $v$ be distinct positive real numbers. (a) Prove by contradiction that \(u+v2\sqrt{

Question

Let $u$ and $v$ be distinct positive real numbers. (a) Prove by contradiction that $u+v>2\sqrt{uv}$. (b) Give a counterexample to show that the strict inequality in part (a) need not hold when the condition $u\ne v$ is removed.

Accepted Answer

$u+v>2\sqrt{uv}$ for distinct positive $u,v$; $u=v=1$ gives equality.

Strict arithmetic-geometric mean inequality

Strict arithmetic-geometric mean inequality

Solution:

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