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.
 A: You can actually do it with Cauchy:
$$
\left(\sum_{cyc}\left(\frac{a}{b+c}\right)^2\right)\left(\sum_{cyc}\left(a(b+c)\right)^2\right)≥\left(a^2+b^2+c^2\right)^2
$$
Thus it is enough to prove:
$$
\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)≥\frac{3}{4}\sum_{cyc}\left(a(b+c)\right)^2=\frac{3}{2}\sum_{cyc}a^2b^2+\frac{3}{2}\sum_{cyc}a^2bc
$$
After little simplification, this reduces to:
$$
\sum_{sym}a^3b≥\frac{3}{2}\sum_{cyc}a^2b^2+\frac{1}{2}\sum_{cyc}a^2bc
$$
Now we can see that:
$$
a^3b+b^3a≥2a^2b^2
$$
And
$$
a^3b+a^3c+ab^3+ac^3≥4a^2bc
$$
By AM-GM. By symmetrically adding these inequalities, we obtain:
$$
2\sum_{sym}a^3b≥4\sum_{cyc}a^2b^2\\
4\sum_{sym}a^3b≥8\sum_{cyc}a^2bc
$$
Which yields directly the inequality remaining to prove.
A: HINT: i think your inequality must be reversed, try $a=1,b=2,c=3$
We can prove it by BW
A: 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:
$$7X^4 - 2S^2X^2 - 3SP^3 \leq 0$$
A: $$\sum_{cyc}\frac{a^2}{(b+c)^2}-\frac{3(a^2+b^2+c^2)}{4(ab+ac+bc)}=$$
$$=\sum_{cyc}\left(\frac{a^2}{(b+c)^2}-\frac{3a^2}{4(ab+ac+bc)}\right)=$$
$$=\sum_{cyc}\frac{a^2(4ab+4ac-2bc-3b^2-3c^2)}{4(ab+ac+bc)(b+c)^2}=$$
$$=\sum_{cyc}\frac{a^2((a-b)(3b+c)-(c-a)(3c+b))}{4(ab+ac+bc)(b+c)^2}=$$
$$=\frac{1}{4(ab+ac+bc)}\sum_{cyc}(a-b)\left(\frac{a^2(3b+c)}{(b+c)^2}-\frac{b^2(3a+c)}{(a+c)^2}\right)\geq0$$
because
 $sign(a-b)=sign(a(3b+c)- b(3a+c))=sign(a-b)=sign((a+c)^2-(b+c)^2).$
Done!
