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I have to solve the following problem by prooving using Contradiction. The problem is:

Show that for any nonnegative real numbers $a$ and $b$ we have: $\frac {a+b}2 ≥ \sqrt {ab}$

(Sadly, i dont even know where to start for this question)

tc216
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destroted
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2 Answers2

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Always to start a proof by contradiction you should assume the negation of the desired result and try to find a contradiction. So let assume that there's $a,b\in\Bbb R_{\ge0}$ s.t. $$\frac{a+b}2<\sqrt{ab}\iff a+b<2\sqrt{ab}$$ so by squaring and developing we find $$a^2+b^2+2ab<4ab\iff a^2+b^2-2ab=(a-b)^2<0\;\text{which's a contradiction}$$ we conclude that the hypothesis is wrong and then its negation: $$\forall a,b\in\Bbb R_{\ge0},\quad \frac{a+b}2\ge\sqrt{ab}$$ is true.

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Hint: Consider $$\left(\sqrt{a}-\sqrt{b}\right)^2$$

This is also known as the arithmetic mean geometric mean inequality.

Henry
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