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!
 A: 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.
A: I will assume your deduction is correct and I will try to prove the inequality to which you have reduced the original inequality.
Consider the sequences
$$(c_i)_{i=1}^{3}:=(\sqrt{bc},\sqrt{ac},\sqrt{ab}),\;\;(d_{i})_{i=1}=(\sqrt{a},\sqrt{b},\sqrt{c})$$
Notice that for any permutation $\sigma:\{1,2,3\}\to\{1,2,3\}$ if
$c_{\sigma(1)}\leq c_{\sigma(2)}\leq c_{\sigma(3)}$ we must have
$d_{\sigma(1)}\geq d_{\sigma(2)}\geq d_{\sigma(3)}$, since $c_{i}d_{i}=\sqrt{abc}$ for all $i$.
Notice also that $b\sqrt{a}+c\sqrt{b}+a\sqrt{c}=c_{3}d_{2}+c_{1}d_{3}+c_{2}d_{1}$, so it follows from
the rearrangement inequality that
$$ b\sqrt{a}+c\sqrt{b}+a\sqrt{c}\geq \sum_{i}c_{i}d_{i}=3\sqrt{abc}. $$
After some obvious manipulations, we conclude that it suffices to show $abc+2-a-b-c$ for any $a,b,c\geq 1$. Replacing $a$ by $1+A$ on both sides this ends up reading as $ABC+\sum AB\geq 0$, which is obviously true for $A,B,C\geq 0$.
A: 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:
$$\frac{ab^2+2}{a+c}+\frac{bc^2+2}{b+a}+\frac{ca^2+2}{b+c} \geq \frac{(b \sqrt a +c \sqrt b+a \sqrt c)^2+18}{2(a+b+c)}$$

A: 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}
A: 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 $$(a-1)(b-1)\geq 0$$
A: 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 .
