Find the largest real $k$ such that: $(a+b+c)^2(ab+bc+ca)\geq k(a^2b^2+b^2c^2+c^2a^2)$

Question

Find the largest real $k$ such that for every non negative real numbers $a,b,c$ : $$(a+b+c)^2(ab+bc+ca)\geq k(a^2b^2+b^2c^2+c^2a^2)$$

I expanded the LHS but the problem got more complicated and no progress...

Answer 1

For $c=0$ and $a=b=1$ we obtain $4\geq k$.

We'll prove that $4$ is a maximal value.

Indeed, by Muirhead we obtain:

$$(a+b+c)^2(ab+ac+bc)=\sum_{cyc}(a^2+2ab)\sum_{cyc}ab=$$ $$=\sum_{cyc}(a^3b+a^3c+a^2bc+2a^2b^2+4a^2bc)=\sum_{cyc}(a^3b+a^3c+2a^2b^2+5a^2bc)\geq$$ $$\geq\sum_{cyc}(4a^2b^2+5a^2bc)\geq4(a^2b^2+a^2c^2+b^2c^2)$$ and we are done!