<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] $$\sum\frac{1}{a+b}+3(ab+bc+ca)\geq\frac{11}{2}$$
$$\sum\frac{(a+b+c)^2}{a+b}+\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$\sum\frac{a^2}{b+c} + 4 +\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$2\sum\frac{a^2(a+b+c)}{b+c}\geq3\sum a^2$$
$$2\sum\frac{a^3}{b+c}\geq\sum a^2$$
Using CSB we get:
$$(\sum\frac{a^3}{b+c})(\sum a(b+c))\geq(\sum a^2)^2 \geq (\sum a^2)(\sum ab)$$
and from that follows the inequality from above.
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