Đáp án:
Giải thích các bước giải:
$ a + b + c = abc ⇔ \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} = 1$
$ \dfrac{1 + \sqrt{1 + a²}}{a} = \dfrac{1}{a} + \sqrt{\dfrac{1}{a²} + 1}$
$ = \dfrac{1}{a} + \sqrt{\dfrac{1}{a²} + \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}}$
$ = \dfrac{1}{a} + \sqrt{(\dfrac{1}{a} + \dfrac{1}{b})(\dfrac{1}{c} + \dfrac{1}{a})}$
$ ≤ \dfrac{1}{a} + \dfrac{1}{2}[(\dfrac{1}{a} + \dfrac{1}{b}) + (\dfrac{1}{c} + \dfrac{1}{a})] = \dfrac{2}{a} + \dfrac{1}{2b} + \dfrac{1}{2c} (1)$
Tương tự:
$ \dfrac{1 + \sqrt{1 + b²}}{b} ≤ \dfrac{2}{b} + \dfrac{1}{2c} + \dfrac{1}{2a} (2)$
$ \dfrac{1 + \sqrt{1 + c²}}{c} ≤ \dfrac{2}{c} + \dfrac{1}{2a} + \dfrac{1}{2b} (3)$
$ (1) + (2) + (3):$
$ \dfrac{1 + \sqrt{1 + a²}}{a} + \dfrac{1 + \sqrt{1 + b²}}{b} + \dfrac{1 + \sqrt{1 + c²}}{c} ≤ 3(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}) = \dfrac{3(ab + bc + ca)}{abc} ≤ \dfrac{(a + b + c)²}{abc} = abc (đpcm)$
Dấu $'=' ⇔ a = b = c ; a + b + c = abc ⇔ a = b = c = \sqrt{3}$
