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I've been asked to prove this inequaltiy $$\dfrac{(x+y)^{1/2}+(x+z)^{1/2}+(y+z)^{1/2}}{\sqrt{(x+y+z)}}<=\sqrt{6}$$

$x, y, z$ are real numbers, $x + y + z\neq 0$

Every time I try the result is nothing.

I appreciate writing proof, but the hint or showing the road(first step) is huge help.

Thank you.

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This does not hold. For $x=y=z=10^{-100}$ the left hand side is $\approx 10^{50}$, for $x=y=z=10^{100}$ it is $\approx 10^{-50}$. – Hagen von Eitzen Feb 11 '13 at 15:05
@amWhy You kept the wrong inequality. – Did Feb 11 '13 at 15:06
Did I edit this correctly? – amWhy Feb 11 '13 at 15:06
Sorry, both are wrong for homogeneity reasons... Maybe a square root in the denominator as well? – Did Feb 11 '13 at 15:08
I was looking at the title: the x, y, z at the end in the body were part of $x, y, z$ are real numbers. Title: "((x+y)^.5 +(x+z)^.5+(y+z)^.5)/(x+y+z)>(6^.5) for every Real x, y, z" – amWhy Feb 11 '13 at 15:08
up vote 2 down vote accepted

Let $a=\sqrt{x+y}$, $b=\sqrt{y+z}$, $c=\sqrt{x+z}$. Then you want to show $$\frac{\sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}}{3}=\frac{a+b+c}3\le \sqrt{\frac{a^2+b^2+c^2}3}=\sqrt{\frac{2(x+y+z)}{3}},$$ which is just the inequality between arithmetic and quadratic mean.

Remark: In the inequality between arithmetic and quadratic mean, we have equality iff $a=b=c$, hence iff $x=y=z$.

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For the record, the inequality to prove might be $$ \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\leqslant\sqrt{6}\cdot\sqrt{x+y+z}, $$ for nonnegative $x$, $y$ and $z$.

To prove this, note that the function $u:a\mapsto\sqrt{a}$ is concave on $a\geqslant0$ hence, for every nonnegative $a$, $b$, $c$, $$ \frac{u(a)+u(b)+u(c)}3\leqslant u\left(\frac{a+b+c}3\right), $$ and apply this inequality to $a=y+z$, $b=z+x$, $c=x+y$.

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you should metion and provide reference to the inequality Jensen's inequality you are using. – achille hui Feb 11 '13 at 15:47
@achillehui Should I? – Did Feb 13 '13 at 17:40
you should at least mention the name of inequality you are using. When OP encounter similar problem, he/she will hopefully know what to do (instead of asking same question here). – achille hui Feb 13 '13 at 18:16

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