For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$ 
For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$

I managed to show this with a 3 instead of a 5 :
The inequality is symmetric in $a$, $b$, $c$ so we can assume $0<c\leq b \leq a$, so $M=\min\{(b-c)^2,(c-a)^2,(a-b)^2\} = \min\{(b-c)^2,(a-b)^2\}$ and $a-c=a-b+b-c \geq 2\min\{(a-b),(b-c)\}$ so $(a-c)^2\geq 4\min\{(a-b),(b-c)\}^2=4M$. Thus we have $6M \leq (b-c)^2+(c-a)^2+(a-b)^2=2(a^2+b^2+c^2-(ab+ac+bc)) \leq 2(a^2+b^2+c^2)$ so $M\leq \frac{a^2+b^2+c^2}{3}$.

Let $a\geq b\geq c$, $b=c+u$ and $a=c+u+v$, where $u$ and $v$ are non-negatives.
Thus, we need to prove that:
$$5\min\{u^2,v^2,(u+v)^2\}\leq c^2+(c+u)^2+(c+u+v)^2$$ or
$$5\min\{u^2,v^2\}\leq3c^2+2c(2u+v)+2u^2+2uv+v^2,$$
for which it's enough to prove that
$$2u^2+2uv+v^2\geq5\min\{u^2,v^2\},$$ which is obvious.
