<s>[INST] For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$ For any real numbers $a,b,c$ show that:
 $$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \leq b \leq c$ because of the symmetry.
$a \leq b \leq c$ implies that $0 \leq b-a \leq c-a$. Since both sides are positive and $y=x^2$ is an increasing function for positive numbers we conclude that $(b-a)^2 \leq (c-a)^2$ and since $(a-b)^2=(b-a)^2$ we obtain $(a-b)^2 \leq (c-a)^2$. 
Therefore:
$\min\{(a-b)^2,(b-c)^2,(c-a)^2\}=$ $$\min\{(b-c)^2,\min\{(a-b)^2,(c-a)^2\}\}=
\min\{(a-b)^2,(b-c)^2\}.$$
So, we have to prove the following inequality instead:
$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
Now two cases can happen:


*

*$$(a-b)^2 \leq (b-c)^2 \implies |a-b| \leq |b-c|=c-b \implies b-c \leq a-b \leq c-b \implies 2b-c \leq a \implies b \leq \frac{a+c}{2}.$$

*$$(b-c)^2 \leq (a-b)^2 \implies |c-b| \leq |a-b|=b-a \implies a-b \leq c-b \leq b-a \implies c \leq 2b-a \implies b \geq \frac{a+c}{2}.$$
So, this all boils down to whether $b$ is greater than the arithmetic mean of $a$ and $b$ are not.
Now I'm stuck. If $a,b,c$ had been assumed to be positive real numbers it would've been a lot easier to go forward from this step. But since we have made no assumptions on the signs of $a,b$ and $c$ I have no idea what I should do next. 
Maybe I shouldn't care about what $\displaystyle \min\{(a-b)^2,(b-c)^2\}$ is equal to and I should continue my argument by dealing with $(a-b)^2$ and $(b-c)^2$ instead. I doubt that using the formula $\displaystyle \min\{x,y\}=\frac{x+y - |x-y|}{2}$ would simplify this any further. Any ideas on how to go further are appreciated.
 [/INST] If we assume that $a\leqslant b\leqslant c$, then it suffices to show$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}.$$


*

*If $c-b\geqslant b-a$, denote $c-b=y,b-a=x$, then $y\geqslant x\geqslant 0$, we just ned to show$$x^2\leqslant \frac{a^2+(a+x)^2+(a+x+y)^2}{2}\iff 3a^2+(4x+2y)a+(y^2+2xy)\geqslant 0.$$Notice that $$\Delta=(4x+2y)^2-4\times 3\times(y^2+2xy)=16x^2-8y-8y^2\leqslant 0,$$since$y\geqslant x\geqslant 0$,so $3a^2+(4x+2y)a+(y^2+2xy)\geqslant 0$ for all $a\in\mathbb{R}$.

*The same method can be applied to solve the condition that $c-b\leqslant b-a$. 

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