Let $ a,b,c $ be nonnegative real numbers such that $ a+b+c=3 $.Prove that
\[ \dfrac{1}{13a^2+(b-c)^2}+\dfrac{1}{13b^2+(c-a)^2}+\dfrac{1}{13c^2+(a-b)^2} \geq \dfrac{3}{13}.\]

Proof:(by scaleye,taiwan)

The inequality is equivalent to \[\frac{1}{13a^2+(b-c)^2}+\frac{1}{13b^2+(c-a)^2}+\frac{1}{13c^2+(a-b)^2} \geq \frac{27}{13(a+b+c)^2}.\]
By Cauchy Schwarz Inequality,
\[
\begin{align*}
\sum\frac{1}{13a^2+(b-c)^2}&=\sum\frac{(a+13b+13c)^2}{\left[13a^2+(b-c)^2\right](a+13b+13c)^2}\\
&\ge \frac{27^2(a+b+c)^2}{\sum\left[13a^2+(b-c)^2\right](a+13b+13c)^2}.
\end{align*}
\]
It suffices to show that
\[13\cdot27(a+b+c)^4 \ge \sum\left[13a^2+(b-c)^2\right](a+13b+13c)^2.\]
After expanding and simplifying, it is equivalent to
\[65\sum(a^3b+ab^3) \ge 122\sum a^2b^2 + 8\sum a^2bc\]
or
\[61\sum ab(a-b)^2 +2\sum(a^3b+ab^3 +a^3c +ac^3 -4a^2bc) \ge 0.\]
Which is true by AM-GM.
Equality holds when $a=b=c=1$  or $a=3$, $b=c=0$ and any  permutations