<s>[INST] 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 $M\leq \frac{a^2+b^2+c^2}{3}$.
 [/INST] Hint: Show that for non-negative real $ a \leq b$,
$$ \min \{ a^2, (b-a)^2 \} \leq \frac{ a^2 + b^2 } { 5 },$$
by considering cases of $ b \geq 2a$ and $ a \leq b < 2a$.
Corollary: The original statement holds.
</s>