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Let $x,y,z >0$. Prove that:

$$\sum_{\text{cyc}}{\sqrt{x^2+xy+y^2}}\geq \sum_{\text{cyc}}{\sqrt{2x^2+xy}} .$$

Thanks for your help :)

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What is $cyc$ ? looks like it holds iff $x^2\leq y^2$ iff... –  Belgi Sep 14 '12 at 10:43
The notation seems still unclear. –  Pedro Tamaroff Sep 14 '12 at 11:32
@downvoter A short explanation why (-1) ? Thanks a lot :) –  Iuli Sep 14 '12 at 12:09
@Iuli I don't know the answer. I voted down, because the notation "cyc" doesn't get explained in the post. I don't understand that notation either. –  Doug Spoonwood Sep 14 '12 at 12:14
@DougSpoonwood If you don't understand something that doesn't mean that something it is not nice, useful . You can ask more information and please review your behavior :) –  Iuli Sep 14 '12 at 12:17

2 Answers 2

up vote 4 down vote accepted

We can write $$ \begin{align} &\sqrt{x^2+xy+y^2}-\sqrt{2x^2+xy}\\ &=\frac{y^2-x^2}{\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}}\\ &=\frac{y+x}{\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}}(y-x)\\ &=\frac{y/x+1}{\sqrt{1+y/x+(y/x)^2}+\sqrt{2+y/x}}(y-x)\tag{1} \end{align} $$ Analysis of the function $$ f(t)=\frac{t+1}{\sqrt{1+t+t^2}+\sqrt{2+t}}\tag{2} $$ shows that it is monotonically increasing. Therefore, $$ (f(y/x)-f(1))(y-x)\ge0\tag{3} $$ Note that $$ \sum_{\mathrm{cyc}}(y-x)=0\tag{4} $$ therefore, $$ \begin{align} &\sum_{\mathrm{cyc}}\left(\sqrt{x^2+xy+y^2}-\sqrt{2x^2+xy}\right)\\ &=\sum_{\mathrm{cyc}}f(y/x)(y-x)\\ &=\sum_{\mathrm{cyc}}(f(y/x)-1)(y-x)\\ &\ge0\tag{5} \end{align} $$ Thus, $$ \sum_{\mathrm{cyc}}\sqrt{x^2+xy+y^2}\ge\sum_{\mathrm{cyc}}\sqrt{2x^2+xy}\tag{6} $$

To see that $f$ is monotonically increasing, let's look at the reciprocal of its square: $$ \begin{align} \frac1{f(t)^2} &=\frac{(t+1)^2+2+2\sqrt{(t+1)^3+1}}{(t+1)^2}\\ &=1+\frac2{(t+1)^2}+2\sqrt{\frac1{t+1}+\frac1{(t+1)^4}}\tag{7} \end{align} $$ and $(7)$ is pretty clearly monotonically decreasing.

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Is there a slick trick showing that $f$ in $(2)$ is monotonic? –  t.b. Sep 14 '12 at 14:37
@t.b.: I originally just plotted it, but I have added a proof. –  robjohn Sep 14 '12 at 15:04
Nice, thanks!${}$ –  t.b. Sep 14 '12 at 15:07

I found a nice idea and I manage to solve it, also this inequality is very interesting.

$$(x-y)^2=x^2-2xy+y^2 \geq 0.$$ We can write this inequality as: $$4x^2+4xy+4y^2 \geq 3x^2+6xy+3y^2 \Leftrightarrow x^2+xy+y^2 \geq \frac{3}{4}(x+y)^2$$ or

$$\sqrt{x^2+xy+y^2} \geq \frac{\sqrt{3}}{2}(x+y).$$

So: $$\sum_{cyc}{\sqrt{x^2+xy+y^2}} \geq \sqrt{3} \cdot (x+y+z)$$

or: $$\left(\sum_{cyc}{\sqrt{x^2+xy+y^2}}\right)^2 \geq 3(x+y+z)^2. \tag{1}$$

Now $\displaystyle \sum_{cyc}{\sqrt{2x^2+xy}}=\sum_{cyc}{\sqrt{x}\cdot \sqrt{2x+y}}$ and we apply Cauchy-Schwarz, so :

$$\left(\sum_{cyc}{\sqrt{x}\cdot\sqrt{2x+y}}\right)^2 \leq (x+y+z)(3(x+y+z))=3(x+y+z)^2. \tag{2}$$

Using relation $(1)$ and relation $(2)$ we obtain the desired result.

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Looks good to me. Maybe you want to write \sum_{\rm cyc} or \sum_{\text{cyc}} to make the cyclic sums look a little better. –  t.b. Sep 14 '12 at 12:31
Looks good to me, too. –  robjohn Sep 14 '12 at 13:12

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