# Prove that $\sqrt{a}\leq\frac{1+a}{2}$

I have a math homework where it's being asked to prove that :

$$\forall a \geq 0,\sqrt{a}\leq\frac{1+a}{2}$$

However, I don't have any idea how I should start this one...

Any idea ?

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Do you know the arithmetic mean- geometric mean inequality? If so, you can apply it directly to $1$ and $a$. –  Dilip Sarwate Sep 4 '12 at 19:59
No I don't already know this one. –  Skydreamer Sep 4 '12 at 20:01
I like the geometric proof as in upload.wikimedia.org/wikipedia/en/7/7c/SemicircleMeans.png –  lhf Sep 4 '12 at 21:34
Seems really nice though I can't understand why the line is $\sqrt{ab}$ long –  Skydreamer Sep 4 '12 at 21:37
@Skydreamer, all triangles with a vertex on the semicircle and basis the diameter are right triangles and the square of the height is the product of the two parts, which follows from Pythagoras's Theorem. –  lhf Sep 4 '12 at 22:32

Try expanding $$(\sqrt a - 1)^2 \geq 0$$

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Well, it was really simple... Sorry for the beginner's question. –  Skydreamer Sep 4 '12 at 20:01
That's OK - I have learned that, no matter how much I think I know about inequalities and how to prove them, what I don't know is far greater. –  marty cohen Sep 4 '12 at 20:04
en.wikipedia.org/wiki/… –  Iuli Sep 4 '12 at 21:15

More generally, let $a,b\geq0$.

$$(a-b)^2 \geq0$$

$$a^2-2ab+b^2 \geq0$$

$$a^2+2ab+b^2 \geq 4ab$$

$$(a+b)^2 \geq 4ab$$

$$\left(\frac{a+b}{2}\right)^2 \geq ab$$

$$\frac{a+b}{2} \geq \sqrt{ab}$$

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