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 :)
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 :)
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!
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.
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.