Is $x+y\geq2\sqrt{xy}$ for every $x,y\geq0$?

Question

According to Wolframalpha, $\forall{x,y\geq0}:x+y\geq2\sqrt{xy}$.

How can I easily prove it?

I think that it has to do with the fact that for a given sum of $x+y$, the value of $xy$ reaches a maximum when $x=y$, thus leading to $x+y=2x=2\sqrt{x^2}=2\sqrt{xy}$.

But I'm not sure how to formulate this proof.

Thank you.

Hint: Try rewritting it as $(a-b)^2 \ge 0$ for suitable $a$ and $b$ — Presage, Nov 16 '20 at 01:06

score 5 · Answer 1 · answered Nov 16 '20 at 01:05

5

$(\sqrt x - \sqrt y)^2\ge0$,

so $x-2\sqrt x \sqrt y +y\ge0$,

so $x+y\ge2\sqrt{xy}.$

answered Nov 16 '20 at 01:05

J. W. Tanner

60,406

I don't understand the reason for the LHS of what you wrote. – goodvibration Nov 16 '20 at 01:06
Oh, the squaring is out of the parenthesis now. – goodvibration Nov 16 '20 at 01:07
Awesome, Tx!!!! – goodvibration Nov 16 '20 at 01:07

score 1 · Answer 2 · answered Nov 16 '20 at 01:07

Since $(x-y)^2$ is nonnegative, we have

$$0 \leq (x-y)^2 = x^2 - 2xy + y^2 = (x^2 + 2xy + y^2) -4xy = (x+y^2)-4xy$$

Hence,

$$ 4xy \leq (x+y)^2 \Longrightarrow xy \leq \frac{(x+y)^2}{4}.$$

Applying square-roots,

$$ \sqrt{xy} \leq \frac{x+y}{2} \leq x+y$$

Is $x+y\geq2\sqrt{xy}$ for every $x,y\geq0$?

2 Answers2