<s>[INST] 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!
 [/INST] By the Buffalo way, WLOG $a=\min\{a,b,c\}$, then set $a=1+k$, $b=1+k+u$ and $c=1+k+v$. Along with the condition $a\ge1$, we get $k$, $u$, $v\ge0$. Plug in and we get
\begin{align}
&(a+b)(b+c)(c+a)(\text{left side}-\text{right side})\\[.6em]
={}&24k^5+40k^4u+40k^4v+120k^4+30k^3u^2+50k^3uv
\\{}+{}&160k^3u+30k^3v^2+160k^3v+168k^3+12k^2u^3\\
{}+{}&18k^2u^2v+90k^2u^2+36k^2uv^2+150k^2uv+168k^2u
\\{}+{}&12k^2v^3+90k^2v^2+168k^2v+72k^2+2ku^4+2ku^3v\\
{}+{}&24ku^3+8ku^2v^2+36ku^2v+72ku^2+14kuv^3+72kuv^2\\
{}+{}&96kuv+48ku+2kv^4+24kv^3+72kv^2+48kv\\
{}+{}&2u^4+2u^3v+12u^3+2u^2v^3+8u^2v^2+9u^2v+16u^2\\
{}+{}&2uv^4+14uv^3+27uv^2+8uv+2v^4+12v^3+16v^2\ge0.
\end{align}
</s>