How to show this inequality $\sum\sqrt{\frac{x}{x+2y+z}}\le 2$

Question

Let $x,y,z,w>0$ show that $$\sqrt{\dfrac{x}{x+2y+z}}+\sqrt{\dfrac{y}{y+2z+w}}+\sqrt{\dfrac{z}{z+2w+x}}+\sqrt{\dfrac{w}{w+2x+y}}\le 2$$

I tried C-S, but without success.

Answer 1

My proof was total wrong. I am sorry.

I think the following reasoning helps.

Let $\frac{x}{x+2y+z}=\frac{a^2}{4}$, $\frac{y}{y+2z+w}=\frac{b^2}{4}$, $\frac{z}{z+2w+x}=\frac{c^2}{4}$ and $\frac{w}{w+2x+y}=\frac{d^2}{4}$, where $a$, $b$, $c$ and $d$ are positives.

Hence, we get that the system

$$ \left\{\begin{matrix} (a^2-4)x+2a^2y+a^2z+0w=0\\ 0x+(b^2-4)y+2b^2z+b^2w=0\\ c^2x+0y+(c^2-4)z+2c^2w=0\\ 2d^2x+d^2y+0z+(d^2-4)w=0 \end{matrix}\right. $$ has infinitely many solutions, which gives $$ \det\left(\begin{matrix} a^2-4&2a^2&a^2&0\\ 0&b^2-4&2b^2&b^2\\ c^2&0&c^2-4&2c^2\\ 2d^2&d^2&0&d^2-4 \end{matrix}\right)=0 $$ or $$a^2b^2+b^2c^2+c^2d^2+d^2a^2+16=4(a^2+b^2+c^2+d^2)+a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2$$ and we need to prove that $$a+b+c+d\leq4.$$ Let $a+b+c+d>4$ and $d=kd'$, where $k>0$ and $a+b+c+d'=4$.

Hence, $k>1$ and $$a^2b^2+b^2c^2+k^2c^2d'^2+k^2d'^2a^2+16=$$ $$=4(a^2+b^2+c^2+k^2d'^2)+a^2b^2c^2+k^2a^2b^2d'^2+k^2a^2c^2d'^2+k^2b^2c^2d'^2$$ or $$a^2b^2+b^2c^2+16-4(a^2+b^2+c^2)-a^2b^2c^2=$$ $$=k^2d'^2(a^2b^2+a^2c^2+b^2c^2-a^2-c^2+4)$$ and since $$4-a^2-c^2=4-\frac{4x}{x+2y+z}-\frac{4z}{z+2w+x}>4\left(1-\frac{x}{x+z}-\frac{z}{z+x}\right)=0,$$ we obtain $$a^2b^2+b^2c^2+16-4(a^2+b^2+c^2)-a^2b^2c^2=$$ $$=k^2d'^2(a^2b^2+a^2c^2+b^2c^2-a^2-c^2+4)>$$ $$>d'^2(a^2b^2+a^2c^2+b^2c^2-a^2-c^2+4),$$ which is contradiction because we'll prove now that $$a^2b^2+b^2c^2+16-4(a^2+b^2+c^2)-a^2b^2c^2\leq$$

$$\leq d'^2(a^2b^2+a^2c^2+b^2c^2-a^2-c^2+4).$$ We'll replace again $d'$ on $d$ and we need to prove that $$a^2b^2+b^2c^2+c^2d^2+d^2a^2+16\leq4(a^2+b^2+c^2+d^2)+a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2$$ or $$a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2+16\leq$$ $$\leq4(a^2+b^2+c^2+d^2)+a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2+a^2c^2+b^2d^2.$$ By AM-GM $a^2c^2+b^2d^2\geq2abcd$.

Id est, it remains to prove the following inequality.

Let $a$, $b$, $c$ and $d$ be positive numbers such that $a+b+c+d=4$. Prove that: $$a^2b^2+b^2c^2+c^2d^2+d^2a^2+a^2c^2+b^2d^2+16\leq$$ $$\leq4(a^2+b^2+c^2+d^2)+a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2+2abcd,$$ which is true, but my proof of this nice inequality is still very ugly.

Answer 2

By C-S:

$(LHS)^2\le \sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}$

$\sum_{cyc}a(b+2c+d) \sum_{cyc}\frac{1}{(a+2b+c)(b+2c+d)}\le 4 \ \ \iff \ \ (a-c)^2(b-d)^2\ge 0$