<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] Hint for $a,b,c,x,y,z\in[1,\infty)$:
Let $x,y,z\in[1,\infty) $ then we have the inequality (tangent line method):
$$f\left(x\right)=\frac{x^{2}y+2}{y+z}\geq f'(1)(x-1)+f(1)$$
remark that again the inequality of the tangent line method is greater than the following inequality :
$$f\left(x,y,z\right)=\frac{\frac{3}{2}\left(\left(2x-1\right)y+2\right)}{x+z+y}$$
Then :
$$f\left(a,b,c\right)+f\left(b,c,a\right)+f\left(c,a,b\right)\geq 4.5$$
Then  $a=b=c=1$ is the minimum we are done .
To show the intermediate inequality we have :
$$\sum_{cyc}\frac{\left(2x-1\right)y\left(2y-z-x\right)}{2\left(x+z\right)\left(x+y+z\right)}\geq 0\tag{I}$$
Because we use :
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}-\frac{4.5}{a+b+c}\geq 0$$
$I$ can be shown using Sum of Square method .
</s>