If $a$, $b$ and $c$ are positives such that $(a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = x$ ($x \ge 9$) then prove that $$\large(a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right) \le x(\sqrt x - 2)^2$$

The equality sign occurs when $a : b : c = m^2 : m : 1$ with $$\large m = \dfrac{\pm \sqrt{x - 2\sqrt x - 3} + \sqrt x - 1}{2}$$, which is very irregular.

Here's what I did.

We have that $$(a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = (ab + bc + ca)\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right) = x$$


$$(a^2 + b^2 + c^2)\left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} \right) = x^2 - 2\left[(ab + bc + ca)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right)^2 + (a + b + c)^2\left(\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} \right)\right] + 4x$$

$$\le x^2 - 4x\sqrt x + 4x = x(\sqrt x- 2)^2$$

Wait, did I just solve this problem by accident? I am saying the truth, I never write down anything at all, I usually type as I think.

Well... I can't calculate when does the equality sign occur. Someone help me.

  • $\begingroup$ I solved your problem. If you want to see my solution show us your trying. $\endgroup$ – Michael Rozenberg Mar 19 at 19:19
  • $\begingroup$ I did, and I accidentally solved the problem in the meantime. $\endgroup$ – Lê Thành Đạt Mar 20 at 13:47

Let $a+b+c=3u$, $ab+ac+bc=v^2$ and $abc=w^3$.

Thus, $a$, $b$ and $c$ are roots of the equation: $$t^3-3ut^2+3v^2t-w^3=0.$$ Now, by your work the equality in your inequality occurs for $$(ab+ac+bc)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=(a+b+c)^2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)$$ or $$\frac{(ab+ac+bc)^3}{a^2b^2c^2}=\frac{(a+b+c)^3}{abc}$$ or $$ab+ac+bc=(a+b+c)\sqrt[3]{abc}$$ or $$v^2=uw.$$ Id est, $x=\frac{9u^2}{w^2}$ and $a$, $b$ and $c$ are roots of the following equation. $$t^3-3ut^2+3uwt-w^3=0$$ or $$(t-w)(t^2+tw+w^2)-3ut(t-w)=0$$ or $$(t-w)(t^2+(w-3u)t+w^2)=0.$$ Can you end it now?


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