Problem 1786

Solution. The inequality is equivalent to
$$a^{2}+b^{2}+c^{2}+\frac{2 a b^{2}}{c}+\frac{2 b c^{2}}{a}+\frac{2 c a^{2}}{b}+\frac{b^{4}}{c^{2}}+\frac{c^{4}}{a^{2}}+\frac{a^{4}}{b^{2}} \geq \frac{12\left(a^{3}+b^{3}+c^{3}\right)}{a+b+c}$$

Using the following identities
$$\begin{aligned} \sum_{c y c} \frac{b^{4}}{c^{2}}-\sum_{c y c} a^{2} & =\sum_{c y c}(b-c)^{2}\left(1+\frac{b}{c}\right)^{2} \\ \sum_{c y c} \frac{a b^{2}}{c}-\sum_{c y c} a b & =\sum_{c y c} \frac{a(b-c)^{2}}{c} \\ \frac{3 \sum_{c y c} a^{3}}{\sum_{c y c} a}-\sum_{c y c} a^{2} & =\frac{\sum_{c y c}(b+c)(b-c)^{2}}{a+b+c} \end{aligned}$$
the inequality can be rewritten as
$$\sum_{c y c}(b-c)^{2}\left(\left(1+\frac{b}{c}\right)^{2}-1-\frac{4(b+c)}{a+b+c}+\frac{2 a}{c}\right) \geq 0$$
or $S_{a}(b-c)^{2}+S_{b}(c-a)^{2}+S_{c}(a-b)^{2} \geq 0$ where
$$\begin{array}{l} S_{a}=\frac{b^{2}}{c^{2}}+\frac{4 a}{a+b+c}+\frac{2(a+b)}{c}-4 \\ S_{b}=\frac{c^{2}}{a^{2}}+\frac{4 b}{a+b+c}+\frac{2(b+c)}{a}-4 \\ S_{c}=\frac{a^{2}}{b^{2}}+\frac{4 c}{a+b+c}+\frac{2(a+c)}{b}-4 \end{array}$$
(i). The first case. $c \geq b \geq a$. Clearly, $S_{b} \geq 0$ and
$$\begin{aligned} S_{a}+S_{b} & =\frac{c^{2}}{a^{2}}+\frac{b^{2}}{c^{2}}+\frac{4(a+b)}{a+b+c}+\frac{2(b+c)}{a}+\frac{2(a+b)}{c}-8 \\ & >\left(\frac{c^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}-4\right)+\left(\frac{2 c}{a}+\frac{2 a}{c}-4\right)+\left(\frac{2 b}{a}-2\right) \geq 0 \end{aligned}$$

Similarly, we have that
$$\begin{aligned} S_{c}+S_{b} & =\frac{c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}+\frac{4(b+c)}{a+b+c}+\frac{2(b+c)}{a}+\frac{2(a+c)}{b}-8 \\ & >\left(\frac{c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}-2\right)+\left(\frac{2 b}{a}+\frac{2 a}{b}-4\right)+\left(\frac{2 c}{a}-2\right) \geq 0 \end{aligned}$$

So the desired result follows because
$$S_{a}(b-c)^{2}+S_{b}(a-c)^{2}+S_{c}(a-b)^{2} \geq\left(S_{a}+S_{b}\right)(b-c)^{2}+\left(S_{b}+S_{c}\right)(a-b)^{2} \geq 0$$
(ii). The second case. $a \geq b \geq c$. Clearly, $S_{a} \geq 1, S_{c} \geq-1+\frac{4 c}{a+b+c}$ and
$$\begin{aligned} S_{a}+2 S_{b} & =\frac{2 c^{2}}{a^{2}}+\frac{b^{2}}{c^{2}}+\frac{8 b+4 a}{a+b+c}+\frac{4(b+c)}{a}+\frac{2(a+b)}{c}-12 \\ & >\left(\frac{8 b+4 a}{a+b+c}-4\right)+\left(\frac{2 b}{a}+\frac{2 a}{c}-4\right)+\left(\frac{2 c}{a}+\frac{2 a}{c}-4\right) \geq 0 \end{aligned}$$

If $2 b \geq a+c$ then we have
$$\begin{aligned} S_{a}+4 S_{b}+S_{c} & \geq \frac{4 c^{2}}{a^{2}}+\frac{b^{2}}{c^{2}}+\frac{16 b+4 a+4 c}{a+b+c}+\frac{8(b+c)}{a}+\frac{2(a+b)}{c}-21 \\ & \geq \frac{4 c^{2}}{a^{2}}+\frac{b^{2}}{c^{2}}+\frac{8(b+c)}{a}+\frac{2(a+b)}{c}-13 \\ & \geq \frac{4 c^{2}}{a^{2}}+\frac{16 c}{a}+\frac{2 a}{c}-10 \geq 2 \sqrt{32}-10 \geq 0 \end{aligned}$$

Consider the cases
If $a+c \leq 2 b$, certainly $2(b-c) \geq a-c$. If $S_{b} \geq 0$ we are done immediately. Otherwise, suppose that $S_{b} \leq 0$, then
$$S_{a}(b-c)^{2}+S_{b}(a-c)^{2}+S_{c}(a-b)^{2} \geq\left(S_{a}+4 S_{b}+S_{c}\right)(b-c)^{2} \geq 0$$

If $a+c \geq 2 b$, we will prove $S_{c}+2 S_{b} \geq 0$, or
$$g(c)=\frac{2 c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}+\frac{8 b+4 c}{a+b+c}+\frac{4(b+c)}{a}+\frac{2(a+c)}{b}-12 \geq 0 .$$

Just notice that $g(c)$ is an increasing function of $c \geq 0$ and $c \geq 2 b-a$, therefore
(.) If $a \geq 2 b$, we have that
$$\begin{array}{c} g(c) \geq g(0)=\frac{a^{2}}{b^{2}}+\frac{8 b}{a+b}+\frac{4 b}{a}+\frac{2 a}{b}-12 \\ =\left(\frac{a+b}{b}+\frac{9 b}{a+b}-6\right)+\left(\frac{a}{b}+\frac{4 b}{a}-4\right)+\left(\frac{a^{2}}{b^{2}}-4\right)+\left(\frac{1}{3}-\frac{b}{a+b}\right)+\frac{2}{3} \geq 0 \end{array}$$
(.) If $a \leq 2 b$, it's easy to infer that
$$g(c) \geq g(2 b-a)=\frac{8 b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}+\frac{4 b}{a}-\frac{4 a}{3 b}-\frac{14}{3} \geq 0$$

We obtain the conclusion because
$$S_{a}(b-c)^{2}+S_{b}(a-c)^{2}+S_{c}(a-b)^{2} \geq\left(S_{a}+2 S_{b}\right)(b-c)^{2}+\left(S_{c}+2 S_{b}\right)(a-b)^{2} \geq 0$$