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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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    @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.2012-09-04

2 Answers 2

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

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    http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means2012-09-04
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More generally, let $a,b\geq0$.

$(a-b)^2 \geq0$

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

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

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

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

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