Problem 1536

12. Let $f(a, b, c)=\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)-\left(a^{2}+b^{2}+c^{2}\right)$, assume $a \leqslant b \leqslant c$, we prove
$$\begin{array}{c} f(a, b, c) \geqslant f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right) \\ (1) \Leftrightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}-\frac{8}{(a+b)^{2}}+\frac{(a+b)^{2}}{2}-\left(a^{2}+b^{2}\right) \geqslant 0 \Leftrightarrow \\ (a-b)^{2}\left(\frac{(a+b)^{2}+2 a b}{a^{2} b^{2}(a+b)^{2}}-\frac{1}{2}\right) \geqslant 0 \end{array}$$

Since $a \leqslant b \leqslant c$, we have $a+b \leqslant 2$, and $2 \sqrt{a b} \leqslant a+b$, so $a b \leqslant 1$, thus $a^{2} b^{2}(a+b)^{2} > 0$. Therefore, inequality (2) holds.

Next, we prove
$$f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right) \geqslant 0, a \leqslant b \leqslant c$$

Let $t=\frac{a+b}{2}$, then
$$\begin{array}{l} 2 t+c=3 \\ f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right) \geqslant 0 \Leftrightarrow \frac{2}{t^{2}}+\frac{1}{c^{2}}-\left(2 t^{2}+c^{2}\right) \geqslant 0 \Leftrightarrow \\ \frac{2}{t^{2}}+\frac{1}{(3-2 t)^{2}}-\left(2 t^{2}+(3-2 t)^{2}\right) \geqslant 0 \Leftrightarrow \\ 2\left(1-t^{4}\right)(3-2 t)^{2}-t^{2}\left[1-(3-2 t)^{4}\right] \geqslant 0 \Leftrightarrow \\ (1-t)\left(3-t-13 t^{2}+23 t^{3}-16 t^{4}+4 t^{5}\right) \geqslant 0 \Leftrightarrow \\ (t-1)^{2}\left(4 t^{4}-12 t^{3}+11 t^{2}-2 t-3\right) \leqslant 0 \end{array}$$

where $0<t \leqslant \frac{3}{2}$.
We need to prove that when $0<t \leqslant \frac{3}{2}$, $4 t^{4}-12 t^{3}+11 t^{2}-2 t-3 \leqslant 0$. Let $g(t)=4 t^{4}-12 t^{3}+11 t^{2}-2 t-3, 0<t \leqslant \frac{3}{2}$. Then
$$g^{\prime}(t)=16 t^{3}-36 t^{2}+22 t-2=2(t-1)\left(8 t^{2}-10 t+1\right)$$

Let $g^{\prime}(t)=0$, we get $t=1, t=\frac{5 \pm \sqrt{17}}{8}$, hence $g(t)$ is decreasing on $\left(0, \frac{5-\sqrt{17}}{8}\right)$, increasing on $\left(\frac{5-\sqrt{17}}{8}, 1\right)$, decreasing on $\left(1, \frac{5+\sqrt{17}}{8}\right)$, and increasing on $\left(\frac{5+\sqrt{17}}{8}, \frac{3}{2}\right)$. It is easy to see that $g(0)<0, g\left(\frac{5-\sqrt{17}}{8}\right)<0, g(1)<0, g\left(\frac{5+\sqrt{17}}{8}\right)<0, g\left(\frac{3}{2}\right)<0$, so $g(t)=4 t^{4}-12 t^{3}+11 t^{2}-2 t-3<0, 0<t \leqslant \frac{3}{2}$. Therefore, inequality (3) holds, and thus the original inequality holds.