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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...2012-09-14
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    @Belgi for example: $$\sum_{cyc}{a}=\sum_{a,b,c}{a}=a+b+c$$ it is a permutation. $a \rightarrow b \rightarrow c \rightarrow a$.2012-09-14
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    The notation seems still unclear.2012-09-14
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    @Iuli From Peter Tamaroff saying the notation seems unclear, it stands to reason that he doesn't understand it. So, logically speaking, how would he explain it?2012-09-14
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    The notation "cyc" could get explained in the post.2012-09-14
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    @DougSpoonwood Please explain you. It seems you know the answer but you don't want to give it. Please give me the answers. I need your answer:)2012-09-14
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    @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.2012-09-14
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    @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 :)2012-09-14
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    To explain "cyc": we've got three variables $x,y,z$ and an expression $f(x,y)$ involving only two of them. Now the cyclic sum simply is $\sum\limits_{\rm cyc} f = f(x,y) + f(y,z) + f(z,x)$. It's probably possible to formalize what it means in all contexts but it's usually clear what is meant.2012-09-14
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    @t.b. So it stands for "cycle". Thanks.2012-09-14
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    @PeterTamaroff: or "cyclic"2012-09-14
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    @t.b.: something like cyclic indices: $$\sum_{i=1}^nf(x_i,x_{i+1},\dots,x_{k+i-1})$$ where $x_{i+n}=x_i$.2012-09-14

3 Answers 3

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We need to prove that $$\sum_{cyc}\left(\sqrt{x^2+xy+y^2}-\sqrt{2x^2+xy}\right)\geq0$$ or $$\sum_{cyc}\frac{y^2-x^2}{\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}}\geq0$$ or $$\sum_{cyc}\left(\frac{y^2-x^2}{\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}}+\frac{x-y}{\sqrt3}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-y)\left(\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}-\sqrt3(x+y)\right)}{\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}}\geq0$$ or $$\sum_{cyc}\frac{(x-y)\left(\sqrt{(x^2+xy+y^2)(2x^2+xy)}-y^2-2xy\right)}{\left(\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}\right)\left(\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}+\sqrt3(x+y)\right)}\geq0$$ or $$\sum_{cyc}\tfrac{(x-y)^2(2x^3+5x^2y+4xy^2+y^3)}{\left(\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}\right)\left(\sqrt{x^2+xy+y^2}+\sqrt{2x^2+xy}+\sqrt3(x+y)\right)\left(\sqrt{(x^2+xy+y^2)(2x^2+xy)}+y^2+2xy\right)}\geq0.$$ Done!

6

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.

5

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.