<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] A solution proceeding from where the OP has stopped:
Let's assume $a=x^2, b=y^2, c=z^2$. We will show:
$$(y^2x+z^2y+x^2z)^2+18 \geq9(x^2+y^2+z^2).$$

To do so, consider the function $$f(x)=(y^2x+z^2y+x^2z)^2+18-9(x^2+y^2+z^2),$$
where $x,y,z \geq 1$. We have:
$$f^{\prime}(x)=2(y^2x+z^2y+x^2z)(y^2+2xz)-18x \\ \geq 2(x+1+x^2)(1+2x)-18x\\ \geq 2(3x) \times 3-18x =0.$$
So, $f(x)$ is an increasing functions. Therefore, we will be done if we prove that $f(1) \geq 0$; in other words, $f(1)=(y^2+z^2y+z)^2+9-9(y^2+z^2) \geq 0.$
Similarly, let's consider the function:
$$g(y)=(y^2+z^2y+z)^2+9-9(y^2+z^2).$$
Then,
$$g^{\prime} (y)=2(y^2+z^2y+z)(2y+z^2)-18y \\ \geq 2(y^2+y+1)(2y+1)-18y\\ \geq 2(3y) \times 3-18y=0;$$
hence $g(y)$ is an increasing function as well. As a result:
$$f(x) \geq f(1)=g(y) \geq g(1)=(1+z^2+z)^2-9z^2 \\ \geq (3z)^2-9z^2=0.$$
We are done.

PS:How did the OP get to that inequality?
By the Cauchy-Schwarz inequality:
$$(\frac{ab^2}{a+c}+\frac{bc^2}{b+a}+\frac{ca^2}{b+c})((a+c)+(b+a)+(a+c))\\ \geq (b \sqrt a +c \sqrt b+a \sqrt c)^2 \\ \implies \frac{ab^2}{a+c}+\frac{bc^2}{b+a}+\frac{ca^2}{b+c} \geq \frac{(b \sqrt a +c \sqrt b+a \sqrt c)^2}{2(a+b+c)}$$
Similarly,
$$(\frac{1}{a+c}+ \frac{1}{b+a}+\frac{1}{b+c})((a+c)+(b+a)+(a+c))\geq (1+1+1)^2 \\ \implies \frac{1}{a+c}+ \frac{1}{b+a}+\frac{1}{b+c} \geq \frac{9}{2(a+b+c)},$$
therefore:
$$\frac{ab^2+2}{a+c}+\frac{bc^2+2}{b+a}+\frac{ca^2+2}{b+c} \geq \frac{(b \sqrt a +c \sqrt b+a \sqrt c)^2+18}{2(a+b+c)}$$

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