Inequality. $\sum{\sqrt{x^2+xy+y^2}}\geq \sum{\sqrt{2x^2+xy}}.$

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 :)

-
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. – Peter 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
show 10 more comments

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.

-
 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.

-
 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