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

Answer 2

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 $$ \sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}}\geq 2\sqrt{2}. $$