How prove $\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \ge \frac{3}{4}+\frac{(a-b)(b-c)(a-c)}{(a+b+c)^3-3abc} $?

Question

Let $a \ge b \ge c >0$ . How can I prove

$$\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \ge \frac{3}{4}+\frac{(a-b)(b-c)(a-c)}{(a+b+c)^3-3abc}. $$ Maybe a simple way?

Answer 1

I think a simplest way here is a full expanding: $$\sum\limits_{cyc}(5a^7b^2+a^7c^2+13a^6b^3+9a^6c^3+18a^5b^4+18a^5c^4+2a^7bc+15a^6b^2c+11a^6c^2b+$$ $$+11a^5b^3c+15a^5c^3b+16a^4b^4c-6a^5b^2c^2-50a^4b^3c^2-50a^4c^3b^2-28a^2b^2c^2)\geq0,$$ which is obviously true.