<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] We have $1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ and thus $$3(ab+bc+ca)=3/2-3/2(a^2+b^2+c^2)$$.
Using this and the fact you found, we find that an equivalent inequality is 
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}-\frac{3}{2}a^2-\frac{3}{2}b^2-\frac{3}{2}a^2\geq1$$
Now $\frac{a}{1-a}\geq a(9/4a+3/4) $ (since $1/(1-a)\geq (9/4)a+3/4$) so we can conclude that
$$
\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}-\frac{3}{2}a^2-\frac{3}{2}b^2-\frac{3}{2}a^2\geq $$
$$\frac{3}{4}(a^2+b^2+c^2)+\frac{3}{4}(a+b+c)\geq$$
$$\frac{3}{4}+\frac{1}{4}=$$
$$1$$
And we're done.
Intuition:
I tried to go for an inequality of the form 
$$\frac{1}{1-a}\geq ax+y$$
in an effort to 'homogenize' the found inequality. Due to the fact that $1/(1-a)$ is convex, we have that any line tangent to its graph on $x\in [0,1)$ would be below it on that region. The reason I chose $$\frac{1}{1-a}\geq \frac{9}{4}a+\frac{3}{4}$$
was that I noticed that the first inequality had equality when $a=b=c=1/3$ and thus I wanted my middle-man inequality to also have equality when $a=1/3$ and thus chose the line tangent to $1/(1-a)$ at $a=1/3$.
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