Alex Baird

2022-07-23

For $a,b,c<1$. Minimize $M=\frac{{a}^{2}\left(1-2b\right)}{b}+\frac{{b}^{2}\left(1-2c\right)}{c}+\frac{{c}^{2}\left(1-2a\right)}{a}$
My Try: From $ab+bc+ca=1⇔\sqrt{3}\le a+b+c<3$
We have: $M=\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-2\left({a}^{2}+{b}^{2}+{c}^{2}\right)$
$\ge \frac{{\left(a+b+c\right)}^{2}}{a+b+c}-2\left({a}^{2}+{b}^{2}+{c}^{2}\right)-4\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)$
$=a+b+c-2\left(a+b+c{\right)}^{2}+4\left(1\right)$
Let $a+b+c=x\left(\sqrt{3}\le x<3\right)$
We find Min of function $y=-2{x}^{2}+x+4$
And ${y}_{Min}=-2+\sqrt{3}$ when $x=\sqrt{3}$
Right or Wrong ?

jbacapzh

Expert

Let $a=b=c=\frac{1}{\sqrt{3}}$. Hence, $M=\sqrt{3}-2$
We'll prove that it's a minimal value.
Indeed, Let $c=min\left\{a,b,c\right\}$. Hence, we need to prove that
$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{2\left({a}^{2}+{b}^{2}+{c}^{2}\right)}{\sqrt{ab+ac+bc}}\ge \left(\sqrt{3}-2\right)\sqrt{ab+ac+bc}$
or
$\sqrt{ab+ac+bc}\left(\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\sqrt{3\left(ab+ac+bc\right)}\right)\ge 2\left({a}^{2}+{b}^{2}+{c}^{2}-ab-ac-bc\right)$
or
$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{a}-a-b+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{{b}^{2}}{a}-c+a+b+c-\sqrt{3\left(ab+ac+bc\right)}\ge$
$\ge 2\left(\left(a-b{\right)}^{2}+\left(c-a\right)\left(c-b\right)\right)$
or
$\frac{\left(a-b{\right)}^{2}\left(a+b\right)}{ab}+\frac{\left(c-a\right)\left({c}^{2}-{b}^{2}\right)}{ac}+a+b+c-\sqrt{3\left(ab+ac+bc\right)}\ge$
$\ge 2\left(\left(a-b{\right)}^{2}+\left(c-a\right)\left(c-b\right)\right)$
or
$\left(a-b{\right)}^{2}\left(\frac{1}{a}+\frac{1}{b}-2\right)+\left(c-a\right)\left(c-b\right)\left(\frac{1}{a}+\frac{b}{ac}-2\right)+a+b+c-\sqrt{3\left(ab+ac+bc\right)}\ge 0,$
which is true because $\frac{1}{a}+\frac{1}{b}-2>0$, $\frac{1}{a}+\frac{b}{ac}-2\ge \frac{1}{a}+\frac{1}{a}-2>0$ and
$a+b+c-\sqrt{3\left(ab+ac+bc\right)}=\frac{\left(a-b{\right)}^{2}+\left(c-a\right)\left(c-b\right)}{a+b+c+\sqrt{3\left(ab+ac+bc\right)}}\ge 0.$
Done!
The inequality
$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{2\left({a}^{2}+{b}^{2}+{c}^{2}\right)}{\sqrt{ab+ac+bc}}\ge \left(\sqrt{3}-2\right)\sqrt{ab+ac+bc}$
is true for all positives $a$, $b$ and $c$, but my proof of the last statement is very ugly.

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