Prove or disprove the inequality for the length of the sides of the triangle $a, b, c$.

Question

Prove or disprove the inequality for the length of the sides of the triangle $a, b, c$ $$\left(a^2+b^2+c^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2\right) \left( \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right) \geq 9.$$ The solution is $$ \left(a^2+b^2+c^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2\right) \left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right) = \\ $$ $$ = \left(a^2+b^2+c^2-a^2+2ab-b^2-b^2+2bc-c^2-c^2+2ac-a^2\right) \left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right) = \\ $$ $$ = \left(2ab+2bc+2ac-a^2-b^2-c^2\right) \left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right) = \dfrac{2ab}{a^2}+\dfrac{2ab}{b^2}+\dfrac{2ab}{c^2} + \\ $$ $$ + \dfrac{2bc}{a^2}+\dfrac{2bc}{b^2}+\dfrac{2bc}{c^2}+\dfrac{2ac}{a^2}+\dfrac{2ac}{b^2}+\dfrac{2ac}{c^2}-1-\dfrac{a^2}{b^2}-\dfrac{a^2}{c^2}-1-\dfrac{b^2}{c^2}-1-\dfrac{c^2}{a^2}-\dfrac{c^2}{b^2}= \\ $$ $$ = \dfrac{2b}{a}+\dfrac{2a}{b}+\dfrac{2ab}{c^2}+\dfrac{2bc}{a^2}+\dfrac{2c}{b}+\dfrac{2b}{c}+\dfrac{2c}{a}+\dfrac{2a}{c}+\dfrac{2ac}{b^2}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}-\dfrac{a^2}{c^2}-\dfrac{c^2}{a^2}-\dfrac{b^2}{c^2}-\dfrac{c^2}{b^2}-3 \stackrel{?}{\geq} 9 $$ $$ \dfrac{2ab}{c^2}+\dfrac{2bc}{a^2}+\dfrac{2ac}{b^2}+\dfrac{2a}{b}+\dfrac{2b}{a}+\dfrac{2b}{c}+\dfrac{2c}{b}+\dfrac{2a}{c}+\dfrac{2c}{a}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}-\dfrac{a^2}{c^2}-\dfrac{c^2}{a^2}-\dfrac{b^2}{c^2}-\dfrac{c^2}{b^2} \stackrel{?}{\geq} 12 \\ $$ $$ \dfrac{2a}{b}+\dfrac{2b}{a} \stackrel{A \geq G}{\geq} 2 \sqrt{\dfrac{2a \cdot 2b}{a \cdot b}}=2 \cdot 2 =4; \ \dfrac{2b}{c}+\dfrac{2c}{b} \stackrel{A \geq G}{\geq} 4; \ \dfrac{2a}{c}+\dfrac{2c}{a} \stackrel{A \geq G}{\geq} 4 \Rightarrow \\ $$ $$ \Rightarrow \dfrac{2a}{b}+\dfrac{2b}{a}+\dfrac{2b}{c}+\dfrac{2c}{b}+\dfrac{2a}{c}+\dfrac{2c}{a} \geq 12; \\ $$ $$ \dfrac{2ab}{c^2}+\dfrac{2bc}{a^2}+\dfrac{2ac}{b^2}-\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}-\dfrac{a^2}{c^2}-\dfrac{c^2}{a^2}-\dfrac{b^2}{c^2}-\dfrac{c^2}{b^2} \stackrel{?}{\geq} 0 \\ $$ $$ \dfrac{2ab}{c^2}+\dfrac{2bc}{a^2}+\dfrac{2ac}{b^2} \stackrel{?}{\geq} \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+\dfrac{a^2}{c^2}+\dfrac{c^2}{a^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{b^2} \\ $$ $$ \dfrac{a^2}{b^2}+\dfrac{b^2}{a^2} \stackrel{A \geq G}{\geq} 2 \sqrt{\dfrac{a^2 \cdot b^2}{b^2 \cdot a^2}} = 2 \sqrt{1}=2; \dfrac{a^2}{c^2}+\dfrac{c^2}{a^2} \stackrel{A \geq G}{\geq} 2; \dfrac{b^2}{c^2}+\dfrac{c^2}{a^2} \stackrel{A \geq G}{\geq} 2 \\ $$ $$ \dfrac{2ab}{c^2}+\dfrac{2bc}{a^2}+\dfrac{2ac}{b^2} \stackrel{?}{\geq} 2+2+2=6 \ | :2 \\ $$ $$ \dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2} \stackrel{?}{\geq} \ | \cdot a^2b^2c^2 \Rightarrow $$ $$ a^3b^3+b^3c^3+a^3c^3 \stackrel{?}{\geq} 3 a^2b^2c^2 \Rightarrow $$ $$ a^3b^3+b^3c^3+a^3c^3 \stackrel{A \geq G}{\geq} 3 \sqrt[3]{a^3b^3 \cdot b^3c^3 \cdot a^3c^3}=3a^2b^2c^2. $$ So the inequality is true. Maybe there exists a little bit shorter way how to prove this inequality? Any hint would help a lot, thanks!

Answer 1

We need to prove that: $$\sum_{cyc}(2ab-a^2)\sum_{cyc}a^2b^2\geq9a^2b^2c^2.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(w^3)\geq0,$ where $f$ is a concave function,

which says that it's enough to prove our inequality for an extremal value of $w^3$, which by $uvw$ (https://artofproblemsolving.com/community/c6h278791) happens in the following cases.

1. Two variables are equal;

2. $w^3\rightarrow0^+$;

3. $a+b-c\rightarrow0^+.$

All these easy to check.

By the way, the case 3 gives more case of the equality occurring, which says that your solution is wrong (you can-not use AM-GM as did you do it. Indeed, after this using you got a wrong inequality already. See a line 8).

Also, your inequality follows from the known Ji Chen's inequality:

Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz\neq0.$ Prove that:

$$(xy+xz+yz)\left(\frac{1}{(x+y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}\right)\geq\frac{9}{4}.$$

Answer 2

Another way.

We need to prove that: $$\sum_{cyc}(2ab-a^2)\sum_{cyc}\frac{1}{a^2}\geq9$$ or $$\sum_{cyc}\left(\frac{\sum\limits_{cyc}(2ab-a^2)}{a^2}-3\right)\geq0$$ or $$\sum_{cyc}\frac{2ab+2ac+2bc-b^2-c^2-4a^2}{a^2}\geq0$$ or $$\sum_{cyc}\frac{(c-a)(2a+b-c)-(a-b)(2a+c-b)}{a^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{2b+c-a}{b^2}-\frac{2a+c-b}{a^2}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2c^2(ab+ac+bc-a^2-b^2)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, since $ab+ac+bc-b^2-c^2\geq0,$ $ab+ac+bc-a^2-c^2=a(b+c-a)+c(b-c)>0$ and $a-c\geq a-b$, we obtain: $$\sum_{cyc}(a-b)^2c^2(ab+ac+bc-a^2-b^2)\geq$$ $$\geq(a-b)^2c^2(ab+ac+bc-a^2-b^2)+(a-c)^2b^2(ab+ac+bc-a^2-c^2)\geq$$ $$\geq(a-b)^2(c^2+c^3a+c^3b-a^2c^2-b^2c^2+b^3a+b^2ac+b^3c-a^2b^2-b^2c^2)=$$ $$=(a-b)^2(bc(b-c)^2+b^2a(b+c-a)+c^2a(b+c-a))\geq0.$$