How prove this in equality with $abc=1$

Question

let $a>$, $b>0$ and $c>0$ such $abc=1$. Show that: $$(ab+bc+ca)\left(\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\right)\ge 3.$$

It see use $a=\dfrac{x}{y},b=\dfrac{y}{z},c=\dfrac{z}{x}$,so we have $$ab+bc+ca=\sum\dfrac{x}{z}$$ $$\sum\dfrac{1}{a^2+2}=\sum\dfrac{y^2}{x^2+2y^2}$$ it engouht to prove $$\sum\dfrac{x}{z}\sum\dfrac{x^2}{x^2+2y^2}\ge 3$$

Answer 1

By C-S $$\sum_{cyc}bc\sum_{cyc}\frac{1}{a^2+2}\geq\left(\sum_{cyc}\sqrt{\frac{bc}{a^2+2}}\right)^2=\left(\sum_{cyc}\sqrt{\frac{1}{a^3+2a}}\right)^2.$$ Let $a=e^x$, $b=e^y$ and $c=e^z$.

Hence, $x+y+z=0$ and it's enough to prove that $\sum\limits_{cyc}f(x)\geq\sqrt3,$ where $$f(x)=\frac{1}{\sqrt{e^{3x}+2e^x}}.$$ But $$f''(x)=\frac{9e^{4x}-4e^{2x}+4}{4\sqrt{e^x(e^{2x}+2)^5}}>0$$ and by Jensen $$\sum\limits_{cyc}f(x)\geq3f\left(\frac{x+y+z}{3}\right)=3f(0)=\sqrt3$$ and we are done!

Answer 2

$uvw$ helps!

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$3v^2\cdot\frac{\sum\limits_{cyc}(a^2b^2+4a^2w^2+4w^4)}{\prod\limits_{cyc}(a^2+2w^2)}\geq3$$ or $$v^2\cdot\frac{9v^4-6uw^3+4(9u^2-6v^2)w^2+12w^4)}{9w^6+2w^2(9v^4-6uw^3)+4(9u^2-6v^2)w^4}\geq1$$ or $f(u)\geq0$, where $$f(u)=3v^6+12u^2v^2w^2-14v^4w^2-2uv^2w^3-8u^2w^4+12v^2w^4-3w^6.$$ But $$f'(u)=24uv^2w^2-2v^2w^3-16uw^4>0,$$ which says that it's enough to prove our inequality for a minimal value of $u$, which happens for equality case of two variables.

Let $b=a$ and $c=\frac{1}{a^2}$.

Hence, we need to prove that $$\left(a^2+\frac{2}{a}\right)\left(\frac{2}{a^2+2}+\frac{a^4}{2a^4+1}\right)\geq3$$ or $$(a-1)^2(a^7+2a^6+3a^5+6a^4-3a^3+2a+4)\geq0,$$ which is obvious.

Done!