How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but had no luck. Thank you.

This is an incomplete answer, but I think the last inequality is (correct) and easier to prove.
We have:
$$\bigg (\frac{a}{b+c} \bigg)^2 + \bigg (\frac{b}{a+c} \bigg)^2 + \bigg (\frac{c}{b+a} \bigg)^2 = \frac{a^4}{a^2(b+c)^2} + \frac{b^4}{b^2(a+c)^2} + \frac{c^4}{c^2(b+a)^2} \geq \frac{(a^2+b^2+c^2)^2}{a^2(b+c)^2 + b^2(a+c)^2 + c^2(b+a)^2}$$
It suffices to show that:
$$\frac{(a^2+b^2+c^2)^2}{a^2(b+c)^2 + b^2(a+c)^2 + c^2(b+a)^2} \geq \frac{3}{4} \frac{a^2+b^2+c^2}{ab+bc+ac}$$
Or equivalently:
$$\frac{(a^2+b^2+c^2)(ab+bc+ac)}{a^2(b+c)^2 + b^2(a+c)^2 + c^2(b+a)^2} \geq \frac{3}{4} (*)$$
The denominator can be written as: $2(a^2b^2 + b^2c^2 + c^2a^2) +2abc(a+b+c) = 2(ab+bc+ac)^2 - 2abc(a+b+c)$
Thus $(*)$ becomes
$$\frac{\bigg ((a+b+c)^2-2(ab+bc+ac) \bigg)(ab+bc+ac)}{(ab+bc+ac)^2 - abc(a+b+c)} \geq \frac{3}{2} (**)$$
Let $S = a+b+c$, $P^3 = abc$, $X^2 = ab+bc+ac$, where $X^2 \leq \frac{S^2}{3}$  Thus we need to show that:
$$7X^4 - 2S^2X^2 - 3SP^3 \leq 0$$
