Show that if $x$ and $y$ are nonnegative reals, then $\frac{x + y}{ 2} ≥ \sqrt{xy}$

Question

Show that if $x$ and $y$ are nonnegative reals, then

$$\frac{x + y}{2} ≥ \sqrt{xy}$$

Can anybody help me?, I'm not quite sure how to prove this.

Answer 1

just note that for $x, y \geqslant 0$: $$ \left(\sqrt{x} - \sqrt{y}\right)^2 \geqslant 0 \implies x + y -2\sqrt{xy} \geqslant 0 \implies \frac{x + y}{2} \geqslant \sqrt{xy} $$

Answer 2

$(\sqrt x - \sqrt y)^2 \ge 0 \iff x-2\sqrt{xy} + y \ge 0 \iff (x+y)/2 \ge \sqrt{xy}$ .

The leftmost inequality is clearly true for $x,y$ non-negative reals.

Answer 3

Since both $\sqrt{x}\geq0$ and $\sqrt{y}\geq0$, then

$$(\sqrt{x}-\sqrt{y})^{2}\geq0$$

Now expand the LHS.

Answer 4

It'll be straightforward from A.M-G.M (as $x, y \geqslant 0$): $$ \text{A.M} \geqslant \text{G.M} \implies \frac{x + y}{2} \geqslant \sqrt{xy} $$

Answer 5

Square both sides, so $$\frac{x^2+2xy+y^2}{4} \ge xy.$$

Multiply by $4$, so $$x^2+2xy+y^2 \ge 4xy.$$

Move terms to one side, so $$x^2-2xy+y^2\ge 0.$$

This factors to $$(x-y)^2\ge 0.$$

Since all real perfect squares are greater than or equal to $0$, the inequality holds.