Alex Baird

Answered

2022-07-23

For $a,b,c<1$. Minimize $M=\frac{{a}^{2}(1-2b)}{b}+\frac{{b}^{2}(1-2c)}{c}+\frac{{c}^{2}(1-2a)}{a}$

My Try: From $ab+bc+ca=1\iff \sqrt{3}\le a+b+c<3$

We have: $M=\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-2({a}^{2}+{b}^{2}+{c}^{2})$

$\ge \frac{{(a+b+c)}^{2}}{a+b+c}-2({a}^{2}+{b}^{2}+{c}^{2})-4(ab+bc+ca)+4(ab+bc+ca)$

$=a+b+c-2(a+b+c{)}^{2}+4(1)$

Let $a+b+c=x(\sqrt{3}\le x<3)$

We find Min of function $y=-2{x}^{2}+x+4$

And ${y}_{Min}=-2+\sqrt{3}$ when $x=\sqrt{3}$

Right or Wrong ?

My Try: From $ab+bc+ca=1\iff \sqrt{3}\le a+b+c<3$

We have: $M=\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-2({a}^{2}+{b}^{2}+{c}^{2})$

$\ge \frac{{(a+b+c)}^{2}}{a+b+c}-2({a}^{2}+{b}^{2}+{c}^{2})-4(ab+bc+ca)+4(ab+bc+ca)$

$=a+b+c-2(a+b+c{)}^{2}+4(1)$

Let $a+b+c=x(\sqrt{3}\le x<3)$

We find Min of function $y=-2{x}^{2}+x+4$

And ${y}_{Min}=-2+\sqrt{3}$ when $x=\sqrt{3}$

Right or Wrong ?

Answer & Explanation

jbacapzh

Expert

2022-07-24Added 18 answers

Let $a=b=c=\frac{1}{\sqrt{3}}$. Hence, $M=\sqrt{3}-2$

We'll prove that it's a minimal value.

Indeed, Let $c=min\{a,b,c\}$. Hence, we need to prove that

$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{2({a}^{2}+{b}^{2}+{c}^{2})}{\sqrt{ab+ac+bc}}\ge (\sqrt{3}-2)\sqrt{ab+ac+bc}$

or

$\sqrt{ab+ac+bc}(\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\sqrt{3(ab+ac+bc)})\ge 2({a}^{2}+{b}^{2}+{c}^{2}-ab-ac-bc)$

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(ab+ac+bc)}\ge $

$\ge 2((a-b{)}^{2}+(c-a)(c-b))$

or

$\frac{(a-b{)}^{2}(a+b)}{ab}+\frac{(c-a)({c}^{2}-{b}^{2})}{ac}+a+b+c-\sqrt{3(ab+ac+bc)}\ge $

$\ge 2((a-b{)}^{2}+(c-a)(c-b))$

or

$(a-b{)}^{2}(\frac{1}{a}+\frac{1}{b}-2)+(c-a)(c-b)(\frac{1}{a}+\frac{b}{ac}-2)+a+b+c-\sqrt{3(ab+ac+bc)}\ge 0,$

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

$a+b+c-\sqrt{3(ab+ac+bc)}=\frac{(a-b{)}^{2}+(c-a)(c-b)}{a+b+c+\sqrt{3(ab+ac+bc)}}\ge 0.$

Done!

The inequality

$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{2({a}^{2}+{b}^{2}+{c}^{2})}{\sqrt{ab+ac+bc}}\ge (\sqrt{3}-2)\sqrt{ab+ac+bc}$

is true for all positives $a$, $b$ and $c$, but my proof of the last statement is very ugly.

We'll prove that it's a minimal value.

Indeed, Let $c=min\{a,b,c\}$. Hence, we need to prove that

$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{2({a}^{2}+{b}^{2}+{c}^{2})}{\sqrt{ab+ac+bc}}\ge (\sqrt{3}-2)\sqrt{ab+ac+bc}$

or

$\sqrt{ab+ac+bc}(\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\sqrt{3(ab+ac+bc)})\ge 2({a}^{2}+{b}^{2}+{c}^{2}-ab-ac-bc)$

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(ab+ac+bc)}\ge $

$\ge 2((a-b{)}^{2}+(c-a)(c-b))$

or

$\frac{(a-b{)}^{2}(a+b)}{ab}+\frac{(c-a)({c}^{2}-{b}^{2})}{ac}+a+b+c-\sqrt{3(ab+ac+bc)}\ge $

$\ge 2((a-b{)}^{2}+(c-a)(c-b))$

or

$(a-b{)}^{2}(\frac{1}{a}+\frac{1}{b}-2)+(c-a)(c-b)(\frac{1}{a}+\frac{b}{ac}-2)+a+b+c-\sqrt{3(ab+ac+bc)}\ge 0,$

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

$a+b+c-\sqrt{3(ab+ac+bc)}=\frac{(a-b{)}^{2}+(c-a)(c-b)}{a+b+c+\sqrt{3(ab+ac+bc)}}\ge 0.$

Done!

The inequality

$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{a}-\frac{2({a}^{2}+{b}^{2}+{c}^{2})}{\sqrt{ab+ac+bc}}\ge (\sqrt{3}-2)\sqrt{ab+ac+bc}$

is true for all positives $a$, $b$ and $c$, but my proof of the last statement is very ugly.

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