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

Question

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 $M\leq \frac{a^2+b^2+c^2}{3}$.

Answer 1

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.

Answer 2

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.