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For $x,y \in \mathbb{R}_{0}^+$ I want to show that $$\sqrt{xy} \leq \frac{2xy}{x+y}$$ only if $x=y$. I think squaring both sides is a equivalence transformation due to $x>0, y>0$ but it didn't get me anywhere.

Can someone give me a hint or a solution?

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  • $\begingroup$ wolframalpha.com/input/… What about this? Originally I was supposed to show that is is always false in case of $x \neq y$. I slightly modified it because I though the case $x=y$ would be true. $\endgroup$ – elfeck Nov 4 '15 at 21:09
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Both $x$ and $y$ are assumed to be positive, so you can square the inequality. This gives $$ xy \le \frac {4x^2y^2}{(x+y)^2} \\ \Longleftrightarrow (x+y)^2 \le 4 xy \\ \Longleftrightarrow (x -y)^2 \le 0 $$ which apparently holds only for $x = y$.

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  • $\begingroup$ Okay I really don't know why I didn't see this. Maybe staring too long at it :/ Thanks! $\endgroup$ – elfeck Nov 4 '15 at 21:17
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After taking reciprocals the inequality is equivalent to $$\sqrt{\frac1x\frac1y}\ge \frac{1/x+1/y}2 $$ whereas the arithmetic-geometric inequality for $\frac1x$ and $\frac 1y$ states $$\sqrt{\frac1x\frac1y}\le \frac{1/x+1/y}2 $$ with equality iff $\frac1x=\frac1y$

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  • $\begingroup$ Thanks for your answer, but I'm not supposed to use the arithmetic-geometric inequality. Also I don't immediately see why taking the reciprocal is an equivalence transformation $\endgroup$ – elfeck Nov 4 '15 at 21:19
  • $\begingroup$ @elfeck Assume you have $x\leq y$ then you multiply it by $1/x$ to obtain $1\leq y/x$ and then with $1/y$ which yields $1/y\leq 1/x$, hence $x\leq y\Leftrightarrow 1/x\geq 1/y$. Just keep the right direction of the inequality in mind. $\endgroup$ – Christian Ivicevic Nov 5 '15 at 12:45
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Since $x$ and $y$ are positive, you can write $u=\sqrt{x}$ and $v=\sqrt{y}$, so the inequality is $$ uv\le\frac{2u^2v^2}{u^2+v^2} $$ that's equivalent to $$ u^2+v^2\le 2uv $$ or $$ (u-v)^2\le 0 $$

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  • $\begingroup$ The numerator should be $2u^2v^2$ maybe? Thats needed to make the next line work $\endgroup$ – charlestoncrabb Nov 4 '15 at 22:29
  • $\begingroup$ @charlestoncrabb Yes, thanks for noting $\endgroup$ – egreg Nov 4 '15 at 22:30
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In general, to prove these types of inequalities you want to work backward (i.e., simplify what you're given) until you get something you know is true, then work forward for your proof. For example, getting rid of the fraction and squaring both sides, then combining terms and factoring gives something true...

Notice that $$(x-y)^2\geq 0.$$ This gives $$0\leq x^2-2xy+y^2,$$ add $4xy$ to both sides, $$4xy\leq x^2+2xy+y^2=(x+y)^2,$$ multiply both sides by $xy$ to get $$(2xy)^2\leq xy(x+y)^2$$ then divide $(x+y)^2$ over and take the square root to get $$\sqrt{xy}\geq\frac{2xy}{x+y}.$$ Of course equality holds when $x=y$.

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