<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] Write given expression as $A+B$ where $$A = \frac{ab^2}{a+c} +\frac{bc^2}{b+a} +\frac{ca^2}{c+b}$$
and
$$B= \frac{2}{a+c} + \frac{2}{b+a} + \frac{2}{c+b}$$
By AM-GM inequality we have
$$ A\geq \frac{3abc}{\sqrt[3]{(a+c)(c+b)(b+a)}}$$
and $$ B\geq \frac{6}{\sqrt[3]{(a+c)(c+b)(b+a)}}$$
Further again by AM-GM we have $$\sqrt[3]{(a+c)(c+b)(b+a)}\leq {2(a+b+c)\over 3}$$
So $$A\geq {9abc\over 2(a+b+c)}\;\;\;{\rm and}\;\;\; B\geq {9\over a+b+c}$$
We are left to check if $${9abc\over 2(a+b+c)} +{9\over a+b+c} \geq {9\over 2}$$ i.e. $$\boxed{abc+2\geq a+b+c}$$ is true.
Rewrite it like $$(ab-1)c+2\geq a+b$$
and notice that linear function $f(c) = (ab-1)c+2$ is increasing so $f(c)\geq f(1)$ and we are left to check if $f(1)\geq a+b$ i.e. if $ab+1\geq a+b$ which is clearly true since $$(a-1)(b-1)\geq 0$$
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