Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$ Question -
Prove that for all non-negative real numbers a,b, c, we have
$$
\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}
$$
My work - 
we may assume that $a b c=1$ 
The problem becomes
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq 2 \sqrt{2}
$$
where $x=a^{3}, y=b^{3}, z=c^{3}$ 
now i did not know where to go from here ...i tried all classic inequalities like chebyshev,re-arangement but none of them working .
can anyone solve this using classic inequalities
any help will be appreciated
thank you

Take
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}
$$
where $x=\frac{a^2}{bc}>0, y=\frac{b^2}{ac}>0, z=\frac{c^2}{ab}>0$.
Note that functions
$$f(w)=\sqrt{\frac{2 w+1}{w+2}}$$ 
are strictly increasing for $w\in[0,\infty)$.
Without the loss of generality assume that $x\geq y\geq z$. Hence
$$\frac{a^2}{bc}\geq \frac{b^2}{ac}\geq \frac{c^2}{ab},\, a\geq b\geq c.$$
Furthermore
$$x\geq\frac{a^2}{c^2}=e^2,\,y\geq\frac{c^2}{a^2}=\frac{1}{e^2},\,z\geq\frac{c^2}{a^2}=\frac{1}{e^2}.$$
Using all of this we get
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq \sqrt{\frac{2 e^2+1}{e^2+2}}+2\sqrt{\frac{2+e^2}{1+2e^2}}.
$$
Note that function on the right hand side has an infimum
$$\inf\limits_{e\rightarrow\infty}\sqrt{\frac{2 e^2+1}{e^2+2}}+2\sqrt{\frac{2+e^2}{1+2e^2}}=2\sqrt{2}$$
So, we get 
$$
\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}\geq 2\sqrt{2}.
$$
