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

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

Answer 1

By C-S twice we obtain: $$\sum_{cyc}\sqrt{\frac{2a^2+bc}{a^2+2bc}}-2\sqrt2=\sum_{cyc}\frac{\sqrt{(2a^2+bc)(a^2+2bc)}}{a^2+2bc}-2\sqrt2\geq$$ $$\geq\sum_{cyc}\frac{\sqrt2(a^2+bc)}{a^2+2bc}-2\sqrt2=\sqrt2\left(\sum_{cyc}\left(\frac{a^2+bc}{a^2+2bc}-\frac{1}{2}\right)-\frac{1}{2}\right)=$$ $$=\sqrt2\left(\sum_{cyc}\frac{a^2}{2(a^2+2bc)}-\frac{1}{2}\right)\geq \sqrt2\left(\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+2bc)}-\frac{1}{2}\right)=0.$$