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.
 A: Note that LHS does not change if you replace $(a,b,c)$ by $(a-t,b-t,c-t)$. Thus we can first minimize RHS with respect to $t$,after we replace $(a,b,c)$ by $(a-t,b-t,c-t)$. It turns out that RHS is minimized when $\displaystyle t = \frac{a+b+c}{3}$, with minimum value being 
$$\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$$
So it suffices to show that $$\min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$$
Without loss of generality, assume that $$a \leq b \leq c$$Then clearly, $\min\{(c-b),(b-a), (c-a)\} = \min\{(c-b),(b-a)\}$, so
$$\min\{(c-b)^2,(b-a)^2, (c-a)^2\} = \min\{(c-b)^2,(b-a)^2\}$$
Moreover,
$$(c-a) = (c-b)+(b-a) \ge 2 \min\{(c-b),(b-a)\}$$ 
and
$$\begin{eqnarray}
\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2) &\ge& \frac{1}{6} (\min\{(c-b),(b-a)\})^2 + (\min\{(c-b),(b-a)\})^2 + (2\min\{(c-b),(b-a)\})^2 \\
&=& \min\{(c-b)^2,(b-a)^2\} \\
&=& \min\{(c-b)^2,(b-a)^2, (c-a)^2\}
\end{eqnarray}$$
A: Note that, since you can change $a,b,c$ to $-a,-b,-c$, you can assume that at least two of the $a,b,c$ are nonnegative. So, first start by supposing that two of your numbers are nonnegative, and after that order them. Therefore, without loss of generality, assume that $a\leq b\leq c$, with $0\leq b\leq c$. Then, for the first case, equivalently you want $$2a^2+2b^2-4ab\leq a^2+b^2+c^2\Rightarrow (a-2b)^2\leq c^2+3b^2.$$ But, $a-2b=a-b-b\leq -b\leq 0$, so $$|a-2b|=2b-a\leq c\Rightarrow (a-2b)^2\leq c^2,$$ so the first case is done. For the second case, $$2b^2+2c^2-4bc\leq a^2+b^2+c^2\Rightarrow(2b-c)^2\leq a^2+3b^2.$$ Now, if $2b-c\leq 0$, since $2b-c\geq a$, you have that $|2b-c|\leq |a|\Rightarrow (2b-c)^2\leq a^2$, and, if $2b-c\geq 0$, you have that $|2b-c|=2b-c\leq 2b-b=b\Rightarrow (2b-c)^2\leq b^2$. So, the second case is also done.
Edit: There is actually no need to consider that two of them are nonnegative. In your first case, equivalently you want $(a-2b)^2\leq c^2+3b^2$. So, if $2b-a\geq 0$, then, since $2b-a\leq c$, you have that $(a-2b)^2\leq c^2$, and, if $2b-a\leq 0$, then $|a-2b|=a-2b\leq-b$, so $|a-2b|^2\leq b^2$. You treat the other case similarly.
A: 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 $c-b\leqslant b-a$. 

