inequality under condition $x+y+z=3$

Question

$x$, $y$ and $z$ being three positive real numbers such that $x+y+z=3$

It is asked to prove that

$$ \dfrac{\sqrt x}{y + z}+ \dfrac{\sqrt y}{x + z} + \dfrac{\sqrt z}{y + x} \ge \dfrac 3 2$$

I tried relating it to Nesbitt but achieved no progress

Thanks for any ideas.

Answer 1

Let $x=a^2,$ $y=b^2$ and $z=c^2$, where $a$, $b$ and $c$ are positives.

Thus, $a^2+b^2+c^2=3$ and $$\sum_{cyc}\frac{\sqrt{x}}{y+z}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{3-a^2}-\frac{1}{2}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-1)(a+3)}{3-a^2}=$$ $$=\frac{1}{2}\sum_{cyc}\left(\frac{(a-1)(a+3)}{3-a^2}-(a^2-1)\right)=\frac{1}{2}\sum_{cyc}\frac{(a-1)^2(a+2)a}{3-a^2}\geq0.$$