Prove $\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b} \geq \frac{9}{2}$ for $a,b,c\geq1$ Prove $\dfrac{ab^2+2}{a+c} +\dfrac{bc^2+2}{b+a} +\dfrac{ca^2+2}{c+b} \geq \dfrac{9}{2}$ for $a,b,c\geq1$.
I tried using the Titu Andreescu form of the Cauchy Schwarz inequality and got to this point:
$\dfrac{(b\sqrt{a}+c\sqrt{b}+a\sqrt{c})^2+18}{a+b+c} \geq 9$.
I don't know what to do next, any help is appreciated.
Thanks!

I will assume your deduction is correct and I will try to prove the inequality to which you have reduced the original inequality.
Consider the sequences
$$(c_i)_{i=1}^{3}:=(\sqrt{bc},\sqrt{ac},\sqrt{ab}),\;\;(d_{i})_{i=1}=(\sqrt{a},\sqrt{b},\sqrt{c})$$
Notice that for any permutation $\sigma:\{1,2,3\}\to\{1,2,3\}$ if
$c_{\sigma(1)}\leq c_{\sigma(2)}\leq c_{\sigma(3)}$ we must have
$d_{\sigma(1)}\geq d_{\sigma(2)}\geq d_{\sigma(3)}$, since $c_{i}d_{i}=\sqrt{abc}$ for all $i$.
Notice also that $b\sqrt{a}+c\sqrt{b}+a\sqrt{c}=c_{3}d_{2}+c_{1}d_{3}+c_{2}d_{1}$, so it follows from
the rearrangement inequality that
$$ b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\geq \sum_{i}c_{i}d_{i}=3\sqrt{abc}. $$
After some obvious manipulations, we conclude that it suffices to show $abc+2-a-b-c$ for any $a,b,c\geq 1$. Replacing $a$ by $1+A$ on both sides this ends up reading as $ABC+\sum AB\geq 0$, which is obviously true for $A,B,C\geq 0$.
