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.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that
$$\frac{(a+b+c)^3\sum\limits_{cyc}(a^2+3ab)}{\prod\limits_{cyc}(a+b)}+3(ab+ac+bc)\geq\frac{11}{2}(a+b+c)^2$$
which is fifth degree, which says that we need to prove a linear inequality of $w^3$,
which says that it's enough to prove the last inequality for an extremal value of $w^3$.
Since $a$, $b$ and $c$ are positive roots of the equation 
$(x-a)(x-b)(x-c)=0$ or $w^3=x^3-3ux^2+3v^2x$,
we see that $w^3$ gets an extremal value for equality case of two variables
or maybe for $w^3\rightarrow0^+$.


*

*$w^3\rightarrow0^+$.


Let $c\rightarrow0^+$ and since the last inequality is homogeneous, we can assume $b=1$,
which gives $2a^4-a^3-a+2\geq0$, which is obvious;


*$b=c=1$, which gives $(a-1)^2(a^2+2a+2)\geq0$.


Done!
