Prove $\frac{a^2}{(a+b)^2} \geqslant \frac{4a^2-b^2-bc+7ca}{4(a+b+c)^2}$

Question

Let $a,\,b,\,c$ are positive numbers. Prove that $$\frac{a^2}{(a+b)^2} \geqslant \frac{4a^2-b^2-bc+7ca}{4(a+b+c)^2}. \quad (*)$$ Note. My proof is use sos. Form $(*)$ we get know problem $$\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c^2}{(c+a)^2} \geqslant \frac{3}{4}.$$

Answer 1

We need to prove that: $$4a^2c^2+(a+b)(a^2-6ab+b^2)c+b^2(a+b)^2\geq0,$$ which is a quadratic inequality of $c$.

If $a^2-6ab+b^2\geq0,$ it's obviously true.

But for $a^2-6ab+b^2\leq0$ it's enough to prove that $$(a^2-6ab+b^2)^2-16a^2b^2\leq0$$ or $$(a^2-2ab+b^2)(a^2-10ab+b^2)\leq0,$$ which is obvious.