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I am having trouble with this problem from my latest homework.

Prove the arithmetic-geometric mean inequality. That is, for two positive real numbers $x,y$, we have $ \sqrt{xy}≤ \frac{x+y}{2} .$ Furthermore, equality occurs if and only if $x = y$.

Any and all help would be appreciated.

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    One way is the following. Let $\sqrt{x} = a$ and $\sqrt{y} = b$. Substitute for x and y in terms of $a$ and $b$. Collect all the terms together on the right side, and factor. Do you recognize a familiar inequality?2011-09-15

5 Answers 5

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Since $x$ and $y$ are positive, we can write them as $x=u^2$, $y=v^2$. Then

$(u-v)^2 \geq 0 \Rightarrow u^2 + v^2 \geq 2uv$

which is precisely it.

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Note that $\frac{x+y}{2}-\sqrt{xy}=\frac{(\sqrt{x}-\sqrt{y})^2}{2}.$

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I am surprised no one has given the following very straightforward algebraic argument: \begin{align} 0\leq(x-y)^2&\Longleftrightarrow 0\leq x^2-2xy+y^2\tag{expand}\\[0.5em] &\Longleftrightarrow 4xy\leq x^2+2xy+y^2\tag{add $4xy$ to both sides}\\[0.5em] &\Longleftrightarrow xy\leq\left(\frac{x+y}{2}\right)^2\tag{div. sides by 4 & factor}\\[0.5em] &\Longleftrightarrow \sqrt{xy}\leq\frac{x+y}{2}.\tag{since $x,y\in\mathbb{R}^+$} \end{align} In regards to equality, notice that $\sqrt{xy}\leq\frac{x+y}{2}\leftrightarrow 2\sqrt{xy}\leq x+y$, and it becomes clear that equality holds if and only if $x=y$. $\blacksquare$

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    Thanks for your extensive reply! I think we do disagree :-) and would value your opinion on EWD1300. Meanwhile I'll try to formulate a question about this topic-- but it might take me a little while.2018-01-19
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$\phantom{Proof without words.........}$ enter image description here

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    Picture from http://math.stackexchange.com/a/922035/589.2016-03-09
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$0\le ({\sqrt x}-{\sqrt y})^{2}$ $0\le x-2{\sqrt {xy}}+y$ $2{\sqrt {xy}}\le x+y$ ${\sqrt {xy}}\le {x+y\over2}$

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    That's Bruno's answer.2016-03-09