Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{9}{2}$$
I only use Nesbitt's inequality. Then we only need that $3(ab+bc+ca)\geq 1$, but it is not correct, because $3(ab+bc+ca)\leq 1$.
Maybe some ideas will be more that grateful.

$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq \frac{11}{2}$$
Using $a+b+c=1$, this is equivalent to resolve:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+3(ab+bc+ca)\geq \frac{5}{2}$$
But we can use MA-MG in the expression:
$$\frac{2}{3}\cdot \frac{a}{b+c}+\frac{3}{2}(ab+ca)=\frac{2a^2}{3a(b+c)}+\frac{3a(b+c)}{2}\geq \sqrt{\frac{2a^2}{3a(b+c)}\cdot \frac{3a(b+c)}{2}}=2a$$
Similar to the above expression:
$$\frac{2}{3}\cdot \frac{b}{c+a}+\frac{3}{2}(ab+bc)\geq 2b$$
$$\frac{2}{3}\cdot \frac{c}{a+b}+\frac{3}{2}(ca+bc)\geq 2c$$
Then we need to prove that
$$\frac{1}{3}\left \lbrace \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right\rbrace \geq \frac{1}{2}$$
And this is Nesbitt's inequality.
