$(a+b)(b+c)(a+c) = 8a^2b^2c^2$ Prove that $\frac{a^5 + b^5}{(b^2 + c^2)(c^2 + a^2)} + \frac{b^5 + c^5}{(c^2 + a^2)(a^2 + b^2)} + \frac{c^5 + a^5}{(a^2 + b^2)(b^2 + c^2)} \geq \frac{3}{2} \cdot (abc)^{\frac{2}{3}}$
And also determine when the similarity condition occurs
My attempt : I tried working backwards $a^2 + b^2 \geq 2ab$ $\frac{a^5 + b^5}{(2bc)(2ac)} + \frac{b^5 + c^5}{(2ac)(2ab)} + \frac{c^5 + a^5}{(2ab)(2bc)} \geq \frac{3}{2}\cdot(abc)^{\frac{2}{3}}$
$\frac{(a^6 + b^6) + (b^6 + c^6) + (c^6 + a^6)}{4a^2b^2c^2} \geq \frac{3}{2} (abc)^{\frac{2}{3}}$
By using AM-GM : $a^6 + b^6 + b^6 + c^6 + a^6 + c^6 \geq 6a^2b^2c^2$
Then simplified to $abc \leq 1$
But this is contradiction because $(a+b)(b+c)(a+c) = 8a^2b^2c^2$ Using AM-GM: $(a+b)(b+c)(a+c) \geq 8abc$ Then subs $(a+b)(b+c)(a+c) = 8a^2b^2c^2$
$8a^2b^2c^2 \geq 8abc$
$abc \geq 1$ . any help would be greatly appreciated
By AM-GM, it suffices to prove that $$3\sqrt[3]{\frac{a^5 + b^5}{(b^2 + c^2)(c^2 +a^2)} \cdot \frac{b^5 + c^5}{(c^2 + a^2)(a^2 +b^2)} \cdot \frac{c^5 + a^5}{(a^2 + b^2)(b^2 +c^2)} } \ge \frac32(abc)^{2/3}$$ or $$\frac{(a^5+b^5)(b^5+c^5)(c^5+a^5)}{(a^2 + b^2)^2(b^2+c^2)^2(c^2+a^2)^2} \ge \frac18 (abc)^2. \tag{1}$$
By Cauchy-Bunyakovsky-Schwarz inequality, we have $$a^5 + b^5 = \frac{(a^5 + b^5)(a + b)}{a + b} \ge \frac{(a^3 + b^3)^2}{a + b} = \frac{(a^3 + b^3)^2(a + b)^2}{(a + b)^3} \ge \frac{(a^2 + b^2)^4}{(a+b)^3}. \tag{2}$$
From (1) and (2), it suffices to prove that $$\frac{(a^2 + b^2)^2(b^2+c^2)^2(c^2+a^2)^2}{(a+b)^3(b+c)^3(c+a)^3} \ge \frac18 (abc)^2.$$
Using $a^2 + b^2 \ge \frac{(a + b)^2}{2}$, it suffices to prove that $$\frac{1}{64}(a + b)(b + c)(c+a) \ge \frac18(abc)^2$$ which is true.
We are done.