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

Question

For $0<a,b,c<1$ and $ab+bc+ca=1$, minimize $$M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$$

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

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

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

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

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

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

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

Right or Wrong ?

Answer 1

Let $a=b=c=\frac{1}{\sqrt3}$. Hence, $M=\sqrt3-2$.

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

Indeed, Let $c=\min\{a,b,c\}$. Hence, we need to prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-\frac{2(a^2+b^2+c^2)}{\sqrt{ab+ac+bc}}\geq(\sqrt3-2)\sqrt{ab+ac+bc}$$ or $$\sqrt{ab+ac+bc}\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-\sqrt{3(ab+ac+bc)}\right)\geq2(a^2+b^2+c^2-ab-ac-bc)$$ or $$\frac{a^2}{b}+\frac{b^2}{a}-a-b+\frac{b^2}{c}+\frac{c^2}{a}-\frac{b^2}{a}-c+a+b+c-\sqrt{3(ab+ac+bc)}\geq$$ $$\geq2((a-b)^2+(c-a)(c-b))$$ or $$\frac{(a-b)^2(a+b)}{ab}+\frac{(c-a)(c^2-b^2)}{ac}+a+b+c-\sqrt{3(ab+ac+bc)}\geq$$ $$\geq2((a-b)^2+(c-a)(c-b))$$ or $$(a-b)^2\left(\frac{1}{a}+\frac{1}{b}-2\right)+(c-a)(c-b)\left(\frac{1}{a}+\frac{b}{ac}-2\right)+a+b+c-\sqrt{3(ab+ac+bc)}\geq0,$$ which is true because $\frac{1}{a}+\frac{1}{b}-2>0$, $\frac{1}{a}+\frac{b}{ac}-2\geq\frac{1}{a}+\frac{1}{a}-2>0$ and $$a+b+c-\sqrt{3(ab+ac+bc)}=\frac{(a-b)^2+(c-a)(c-b)}{a+b+c+\sqrt{3(ab+ac+bc)}}\geq0.$$ Done!

The inequality $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-\frac{2(a^2+b^2+c^2)}{\sqrt{ab+ac+bc}}\geq(\sqrt3-2)\sqrt{ab+ac+bc}$$ is true for all positives $a$, $b$ and $c$, but my proof of the last statement is very ugly.

Answer 2

We will prove that $$M=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-2(a^2+b^2+c^2)\geq -2+\sqrt{3} (*)$$ or equivalently $$\left ( \frac{a^2}{b}-2a+b \right )+\left ( \frac{b^2}{c}-2b+c \right )+\left ( \frac{c^2}{a}-2c+a \right )+(a+b+c-\sqrt{3})-2(a^2+b^2+c^2-1)\geq 0.$$

We have $$\sum_{cyc} \left (\frac{a^2}{b}-2a+b\right)=\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}$$ $$a+b+c-\sqrt{3}=\frac{(a+b+c)^2-3}{a+b+c+\sqrt{3}}=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2(a+b+c+\sqrt{3})}$$ $$2(a^2+b^2+c^2-1)=2(a^2+b^2+c^2-ab-bc-ac)=(a-b)^2+(b-c)^2+(c-a)^2.$$

Hence $$(*) \Leftrightarrow (a-b)^2\left [ \frac{1}{b}+\frac{1}{2(a+b+c+\sqrt{3})}-1 \right ]+(b-c)^2\left [ \frac{1}{c}+\frac{1}{2(a+b+c+\sqrt{3})}-1 \right ]+(c-a)^2\left [ \frac{1}{a}+\frac{1}{2(a+b+c+\sqrt{3})}-1 \right ]\geq 0 $$

This true because $\frac{1}{a};\frac{1}{b};\frac{1}{c}>1$ when $a=b=c=\frac{1}{\sqrt 3}$