Math Olympiad exercise $\frac{x}{\sqrt{3x+2y+z}}+\frac{y}{\sqrt{3y+2z+x}}+\frac{z}{\sqrt{3z+2x+y}}\leq\sqrt{\frac{x+y+z}{2}}$

Question

I would like to prove the inequality

$$\frac{x}{\sqrt{3x+2y+z}}+\frac{y}{\sqrt{3y+2z+x}}+\frac{z}{\sqrt{3z+2x+y}}\leq\sqrt{\frac{x+y+z}{2}},$$

where $x,y,z>0$. What I did so far is the following:

The inequality is homogeneous, thus we can assume $x+y+z=1$. When we square both sides of the inequality and use Cauchy-Schwartz on the left-hand side, it suffices to show:

$$\frac{x}{3x+2y+z}+\frac{y}{3y+2z+x}+\frac{z}{3z+2x+y}\leq \frac{1}{2}.$$ But I couldn't make any real progress from there. For example, we can write the left-hand side, using $x+y+z=1$, as

$$\frac{x}{2+x-z}+\frac{y}{2+y-x}+\frac{z}{2+z-y}.$$

This seems somehow simpler, but how can I proceed?

Answer 1

Now, we can use C-S again.

We need to prove that $$\sum_{cyc}\left(\frac{x}{3x+2y+z}-\frac{1}{3}\right)\leq\frac{1}{2}-1$$ or $$\sum_{cyc}\frac{2y+z}{3x+2y+z}\geq\frac{3}{2}$$ and $$\sum_{cyc}\frac{2y+z}{3x+2y+z}=\sum_{cyc}\frac{(2y+z)^2}{(2y+z)(3x+2y+z)}\geq$$ $$\geq\frac{9(x+y+z)^2}{\sum\limits_{cyc}(2y+z)(3x+2y+z)}=\frac{9(x+y+z)^2}{\sum\limits_{cyc}(5x^2+13xy)}\geq$$ $$\geq\frac{9(x+y+z)^2}{\sum\limits_{cyc}(6x^2+12xy)}=\frac{3}{2}.$$