Welcome to our community

Be a part of something great, join today!

Re: Problem Of The Week # 334 (11 Dec 2018) – Alternative solution [Olinguito]

Olinguito

Well-known member
Apr 22, 2018
251
Re: Problem Of The Week # 334 (11 Dec 2018) – Alternative solution [Olinguito]

I’d just like to comment that my solution is different from Opalg ’s: I used the Cauchy–Schwarz inequality.

Let $a,b$ and $c$ be positive reals such that $a^2+b^2+c^2=\sqrt{ab+bc+ca}−14$.

Determine the values of $a,b$ and $c$.
By the Cauchy–Schwarz inequality,
$$ab+bc+ca\le \sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2+a^2}=a^2+b^2+c^2=\sqrt{ab+bc+ca}-\frac14$$
$\implies\ \left(\sqrt{ab+bc+ca}-\dfrac12\right)^2\ \le\ 0$

$\implies\ \sqrt{ab+bc+ca}\ =\ \dfrac12$

$\implies\ a^2+b^2+c^2=\sqrt{ab+bc+ca}-\dfrac14=\dfrac12-\dfrac14=\dfrac14=ab+bc+ca$.

So the C–S inequality is actually equality. The condition for this is that
$$\frac ab=\frac bc = \frac ca$$
from which it follows that $a=b=c$. Hence
$$\frac 14=a^2+b^2+c^2=3a^2$$
$\implies\ \boxed{a=b=c=\dfrac1{\sqrt{12}}}$.