MHB Can You Solve for a, b, and c Given This Challenging Equation?

[anemone]

Here is this week's POTW:

-----

Let $a,\,b$ and $c$ be positive reals such that $a^2+b^2+c^2=\sqrt{ab+bc+ca}-\dfrac{1}{4}.$

Determine the values of $a,\,b $ and $c$.

-----

Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution!(Cool)

1. Olinguito
2. Opalg

Partial credit goes to Ackbach for getting the correct solution but not proving the solution is unique.

Solution from Opalg:

Spoiler

Let $f(a,b,c) = a^2 + b^2 + c^2 - \sqrt{bc+ca+ab}$.

Since $0\leqslant (b-c)^2 + (c-a)^2 + (a-b)^2 = 2(a^2 + b^2 + c^2) - 2(bc+ca+ab)$ (with equality only if $a=b=c$), it follows that $bc+ca+ab \leqslant a^2 + b^2 + c^2$ and therefore $f(a,b,c) \geqslant a^2 + b^2 + c^2 - \sqrt{a^2+b^2+c^2}$ (with equality only if $a=b=c$). But the function $x - \sqrt x$ has minimum value $-\frac14$, attained only at the point $x = \frac14$. Therefore $f(a,b,c) \geqslant -\frac14$, with equality only if $a=b=c$ and $a^2+b^2+c^2 = \frac14$. But if $3a^2 = \frac14$ then $a = \frac1{\sqrt{12}}$.

So the only solution of the equation $f(a,b,c) = -\frac14$ is $a=b=c=\frac1{\sqrt{12}}$.