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.
On
$$\sum\frac{1}{a+b}+3(ab+bc+ca)\geq\frac{11}{2}$$
$$\sum\frac{(a+b+c)^2}{a+b}+\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$\sum\frac{a^2}{b+c} + 4 +\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$2\sum\frac{a^2(a+b+c)}{b+c}\geq3\sum a^2$$
$$2\sum\frac{a^3}{b+c}\geq\sum a^2$$
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.
On
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 $a=1/3$.
On
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 $w^3\rightarrow0^+$.
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;
Done!
On
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 $x \in (0, 1)$, $$f(x) = \frac1{1-x}-\frac32x^2-\frac43 -\frac54\left(x-\frac13\right) = \frac{(1+2x)(1-3x)^2}{12(1-x)}\geqslant 0$$
On
Variational Approach
Let $x=a+b$, $y=b+c$, and $z=c+a$. Then $x+y+z=2$ and we are minimizing $$ \frac1x+\frac1y+\frac1z+3\left(1-\frac{x^2+y^2+z^2}2\right)\tag{1} $$ Furthermore, since $a+b+c=1$ and $a,b,c\ge0$, we have that $0\le x,y,z\le1$.
Since $x+y+z=2$, we get the restriction $$ \delta x+\delta y+\delta z=0\tag{2} $$ Taking the variation of $(1)$ gives $$ \left(\frac1{x^2}+3x\right)\delta x+\left(\frac1{y^2}+3y\right)\delta y+\left(\frac1{z^2}+3z\right)\delta z=0\tag{3} $$ Thus, to minimize $(1)$, we need to have $(3)$ whenever we have $(2)$. This means that $$ \frac1{x^2}+3x=\frac1{y^2}+3y=\frac1{z^2}+3z\tag{4} $$
Case $\boldsymbol{1}$: If $x\ne y$ and $y\ne z$ and $z\ne x$, then $(4)$ implies $$ \frac{x+y}{x^2y^2}=\frac{y+z}{y^2z^2}=\frac{z+x}{z^2x^2}=3\tag{5} $$ which implies that $x=y=z$.
Case $\boldsymbol{2}$: If $x=y\ne z$, then $2x+z=2$ and $$ \frac{z+x}{z^2x^2}=3\implies12x^4-24x^3+12x^2+x-2=0\tag{6} $$ Sturm's Theorem shows that there are no solutions to $(6)$ in $[0,1]$.
Therefore, $(1)$ is minimized when $x=y=z=\frac23$ and thus, $a=b=c=\frac13$. This means that $$ \frac1{a+b}+\frac1{b+c}+\frac1{c+a}+3(ab+bc+ca)\ge\frac{11}2\tag{7} $$
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq \frac{11}{2}$$ 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.