Prove that, for all positive integers $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$

Question

Help I'm really stuck on a proof question:

Prove that, for all positive integers for $x$ and $y$, $\sqrt{ xy} \leq \frac{x + y}{2}$

Thanks!

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score 4 · Answer 1 · answered Oct 08 '17 at 21:08

4

Start with $(\sqrt{x} - \sqrt{y})^2 \geq 0$.

answered Oct 08 '17 at 21:08

Joppy

score 1 · Accepted Answer · answered Oct 08 '17 at 21:09

1

\begin{eqnarray*} (\sqrt{x}- \sqrt{y})^2 \geq 0 \end{eqnarray*} and now do a little bit of algebra.

answered Oct 08 '17 at 21:09

Oh I see! So you expand the brackets and rearrange? – AbiH Oct 08 '17 at 21:10
That's right. Move the $-2 \sqrt{xy}$ to the RHS and divide by $2$. – Donald Splutterwit Oct 08 '17 at 21:12

2 Answers2