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!

Update: Add proof 2 without using Holder inequality.
Proof 2:
We have
\begin{align*}
 &\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b}\\[6pt]
 ={}& \frac{ab^2}{a+c} +\frac{bc^2}{b+a} +\frac{ca^2}{c+b} + \frac{2}{a+c} + \frac{2}{b+a} + \frac{2}{c+b}\\[6pt]
 \ge{}&3\sqrt[3]{\frac{ab^2}{a+c} \cdot\frac{bc^2}{b+a} \cdot\frac{ca^2}{c+b}} + 3\sqrt[3]{\frac{2}{a+c} \cdot \frac{2}{b+a} \cdot \frac{2}{c+b}} \tag{1}\\[6pt]
 ={}& \frac{3abc + 6}{\sqrt[3]{(a+b)(b+c)(c+a)}}\\[6pt]
 \ge{}& \frac{3abc + 6}{\frac{a+b + b+c + c+a}{3}} \tag{2}\\[6pt]
 ={}& \frac{9abc + 18}{2(a + b + c)}\\[6pt]
 ={}& \frac92 \cdot \frac{abc + 2}{a + b + c}\\
 \ge{}& \frac92.\tag{3}
\end{align*}
Explanations:
(1): AM-GM.
(2): AM-GM.
(3): $$abc + 2 - a - b - c$$
$$= 
(a-1)(b-1)(c-1) + (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1) \ge 0.$$
We are done.

Proof 1:
We have
\begin{align*}
 &\frac{ab^2+2}{a+c} +\frac{bc^2+2}{b+a} +\frac{ca^2+2}{c+b}\\[6pt]
 \ge{}& 3\sqrt[3]{\frac{ab^2+2}{a+c} \cdot \frac{bc^2+2}{b+a} \cdot \frac{ca^2+2}{c+b}} \tag{1}\\[6pt]
 \ge{}& 3\sqrt[3]{\frac{(abc + 2)^3}{\left(\frac{a + c + b + a + c + b}{3}\right)^3}} \tag{2}\\[6pt]
 ={}& \frac92 \cdot \frac{abc + 2}{a + b + c}\\
 \ge{}& \frac92. \tag{3}
\end{align*}
Explanations:
(1): AM-GM.
(2): $(ab^2 + 2)(bc^2 + 2)(ca^2 + 2) \ge (\sqrt[3]{ab^2\cdot bc^2 \cdot ca^2} + 2)^3$ by Holder's inequality;
$(a+c)(b+a)(c+b) \le \left(\frac{a + c + b + a + c + b}{3}\right)^3$ by AM-GM.
(3): $$abc + 2 - a - b - c$$
$$= 
(a-1)(b-1)(c-1) + (a-1)(b-1) + (b-1)(c-1) + (c-1)(a-1) \ge 0.$$
We are done.
