<s>[INST] 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.
 [/INST] Hint: The given inequality is equivalent to 
$$\sum_{cyc} \frac1{1-a} - \frac32\sum_{cyc}a^2 \ge 4 $$
which follows from noting that for $x \in (0, 1)$, 
$$f(x) = \frac1{1-x}-\frac32x^2-\frac43 -\frac54\left(x-\frac13\right) = \frac{(1+2x)(1-3x)^2}{12(1-x)}\geqslant 0$$
</s>