Problem: Let $a,b,c>0: \sqrt{a^4+8a}+\sqrt{b^4+8b}+\sqrt{c^4+8c}=3(a^2+b^2+c^2).$ Prove that: $$\frac{a^4+2a+b(2a^2+b)}{a^2(b^2+2a)}+\frac{b^4+2b+c(2b^2+c)}{b^2(c^2+2b)}+\frac{c^4+2c+a(2c^2+a)}{c^2(a^2+2c)}\ge 6.$$
My attempt: The problem is equivalent to: $$\sum_{cyc}{\left(\frac{1}{a^2}+\frac{a^2+2b}{b^2+2a}\right)}\ge6$$ It is complicated problem at first glance. I think the only ostable is using condition. Anyone help me end the rest part? Thanks!
Since $$\sum_{cyc}3a^2=\sum_{cyc}\sqrt{a^4+8a}\leq\sum_{cyc}(a^2+2),$$ we obtain $$a^2+b^2+c^2\leq3.$$ Now, by your work and C-S we obtain: $$\sum_{cyc}\frac{a^2+2b}{b^2+2a}\geq\frac{\left(\sum\limits_{cyc}(a^2+2a)\right)^2}{\sum\limits_{cyc}(a^2+2b)(b^2+2a)}=\frac{\left(\sum\limits_{cyc}(a^2+2a)\right)^2}{\sum\limits_{cyc}(a^2b^2+4a^3+4ab)}.$$ Let $a=kx,$ $b=ky$ and $c=kz$, where $k>0$ and $x^2+y^2+z^2=3$.
Thus, $0<k\leq1$ and we need to prove that: $$\frac{1}{k^2}\sum_{cyc}\frac{1}{x^2}+\frac{\left(3k+2(x+y+z)\right)^2}{\sum\limits_{cyc}(k^2x^2y^2+4kx^3+4xy)}\geq6$$ or $f(k)\geq0,$ where $$f(k)=\frac{1}{k^2}\sum_{cyc}\frac{1}{x^2}+\frac{\left(3k+2(x+y+z)\right)^2}{\sum\limits_{cyc}(k^2x^2y^2+4kx^3+4xy)}-6.$$ But easy to see that $f'(k)<0,$ which says that it's enough to prove $f(k)\geq0$ for $k=1$ and we need to prove that: $$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{\left(3+2(x+y+z)\right)^2}{\sum\limits_{cyc}(x^2y^2+4x^3+4xy)}\geq6,$$ where $x$, $y$ and $z$ are positives such that $x^2+y^2+z^2=3.$
Now, we'll prove that: $$\sum\limits_{cyc}(x^2y^2+4x^3+4xy)\leq27.$$ Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition $x^2+y^2+z^2=3$ does not depend on $w^3$, which says that we need to prove a linear inequality of $w^3$.
But a linear function gets a maximal value for an extremal value of $w^3$, which by $uvw$ happens in the following gases.
Let $z\rightarrow0^+$ and $x^2+y^2=3$.
Thus, $$\sum\limits_{cyc}(x^2y^2+4x^3+4xy)=x^2y^2+4(x^3+y^3)+4xy=$$ $$=x^2y^2+4(x+y)(3-xy)+4xy\leq x^2y^2+4\sqrt{3}(3-xy)+4xy<27,$$ where the last inequality is true for $0<xy\leq\frac{x^2+y^2}{2}=\frac{3}{2}.$
Let $y=x$ and $z=\sqrt{3-2x^2},$ where $0<x<\sqrt{1.5}$.
Thus, we need to prove that: $$x^4+2x^2(3-2x^2)+4\left(2x^3+\left(\sqrt{3-2x^2}\right)^3\right)+4\left(x^2+2x\sqrt{3-2x^2}\right)\leq27$$ or $$3x^4-8x^3-10x^2+27\geq(12+8x-8x^2)\sqrt{3-2x^2}$$ and since $3x^4-8x^3-10x^2+27>0$ for $0<x<\sqrt{1.5},$ it's enough to prove that: $$(3x^4-8x^3-10x^2+27)^2\geq16(3+2x-2x^2)^2(3-2x^2)$$ or $$(x-1)^2(3x^6-10x^5+21x^4+20x^3-43x^2+6x+99)\geq0,$$ which is obvious.
Id est, it remains to prove that: $$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{(3+2(x+y+z))^2}{27}\geq6,$$ which is true by AM-GM: $$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{(3+2(x+y+z))^2}{27}=$$ $$=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{\left(\sqrt{3(x^2+y^2+z^2)}+2(x+y+z)\right)^2}{27}\geq$$ $$\geq\frac{3}{\sqrt[3]{x^2y^2z^2}}+\frac{\left(\sqrt{3\cdot3\sqrt[3]{x^2y^2z^2}}+6\sqrt[3]{xyz}\right)^2}{27}=3\left(\frac{1}{w^2}+w^2\right)\geq6.$$
Alternative proof.
The OP is equivalent to$$\frac{a^2+2b}{b^2+2a}+\frac{b^2+2c}{c^2+2b}+\frac{c^2+2a}{a^2+2c}\ge 6-\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2}.$$ By given hypothesis, $2(a^2+b^2+c^2)=\sum\limits_{cyc}\left(\sqrt{a^4+8a}-a^2\right)=\sum\limits_{cyc}\dfrac{8a}{\sqrt{a^4+8a}+a^2}.$ Hence$$a^2+b^2+c^2=\sum\limits_{cyc}\dfrac{4a}{\sqrt{a^4+8a}+a^2}.\tag{1}$$ Now, we can continue by coming up an idea ispired by Michael Rozenberg's CBS using . Indeed, \begin{align*} (c^2+2a)\left(\sqrt{a^4+8a}+a^2\right)&=a^2c^2+2a^3+\sqrt{[(2a-c^2)^2+8ac^2][a^4+8a]}\\&\ge a^2c^2+2a^3+ (2a-c^2)a^2+8ac\\&=4a(a^2+2c) \end{align*} It implies that $$\frac{c^2+2a}{a^2+2c}\ge \dfrac{4a}{\sqrt{a^4+8a}+a^2}.\tag{2}$$ Sum cyclically on $(2)$ and replace $(1)$ we obtain$$\frac{a^2+2b}{b^2+2a}+\frac{b^2+2c}{c^2+2b}+\frac{c^2+2a}{a^2+2c}\ge a^2+b^2+c^2.$$ It remains to prove $$a^2+b^2+c^2 \ge 6-\frac{1}{a^2}-\frac{1}{b^2}-\frac{1}{c^2},$$ or $$a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \ge 6,$$ $$\iff \left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2\ge 0$$We end the proof here. Equality holds at $a=b=c=1.$
