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

Question

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 :)

Answer 1

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, $$ \sum_{\mathrm{cyc}}\sqrt{x^2+xy+y^2}\ge\sum_{\mathrm{cyc}}\sqrt{2x^2+xy}\tag{6} $$

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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.

Answer 2

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.