Математикадан «Туймаада» олимпиадасы. Жоғары лига. 2014 жыл

Оң $a$, $b$, $c$ сандары $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3$ шартын қанағаттандырса, келесі теңсіздікті дәлелдеңіз: $\dfrac{1}{\sqrt{{{a}^{3}}+1}}+\dfrac{1}{\sqrt{{{b}^{3}}+1}}+\dfrac{1}{\sqrt{{{c}^{3}}+1}}\le \dfrac{3}{\sqrt{2}}$. ( Н. Александров )

Комментарий/решение

$\frac{1}{\sqrt{a^3+1}}\le \frac{\sqrt{a+1}}{a^2+1} \le \frac{\sqrt{a+1}}{2a} = \frac{2\sqrt{(a+1)2}}{4\sqrt{2}a} \le \frac{(a+1)+2}{4\sqrt{2}a} = \frac{1}{4\sqrt{2}} +\frac{3}{4\sqrt{2}a}$

$$\frac{1}{\sqrt{a^3+1}}+\frac{1}{\sqrt{b^3+1}}+\frac{1}{\sqrt{c^3+1}}\leq\frac{3}{\sqrt{2}}\Rightarrow$$

$$\Rightarrow \mathbb{B}=(\frac{1}{\sqrt{a^3+1}}+\frac{1}{\sqrt{b^3+1}}+\frac{1}{\sqrt{c^3+1}})^2\leq \frac{9}{2}\Rightarrow$$

$$\Rightarrow (\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1})(\frac{1}{a^2-a+1}+\frac{1}{a^2-a+1}+\frac{1}{a^2-a+1})\geq (\frac{1}{\sqrt{a^3+1}}+\frac{1}{\sqrt{b^3+1}}+\frac{1}{\sqrt{c^3+1}})^2$$

$$\frac{1}{a^2-a+1}\leq \frac{1}{a}$$ $$\frac{1}{a+1}\leq \frac{1+\frac{1}{a}}{4}\Rightarrow$$

$$\Rightarrow \mathbb{B}\leq \frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{9}{2}$$

$x=\frac{1}{a}; y=\frac{1}{b}; z=\frac{1}{c} \implies x+y+z=3$

$\sum \frac{1}{\sqrt{a^3+1}} = \sum \sqrt{\frac{x^3}{x^3+1}} \leq \frac{3}{\sqrt{2}}$

$(x-1)^2\times (9x^3+24x^2+8x+1) \geq 0 \iff \sqrt{\frac{x^3}{x^3+1}} \leq \frac{3x+1}{\sqrt{2^5}} \implies \sum \sqrt{\frac{x^3}{x^3+1}} \leq \sum \frac{3x+1}{\sqrt{2^5}} = \frac{3}{\sqrt{2}}$

$x+y\ge 2\sqrt{xy} \ \ $ және орташа дәрежелік теңсіздіктерінен:

$\sqrt{\frac{2}{a^3+1}}+\sqrt{\frac{2}{b^3+1}}+\sqrt{\frac{2}{c^3+1}} \le \sqrt{\frac{2}{2\sqrt{a^3}}}+\sqrt{\frac{2}{2\sqrt{b^3}}}+\sqrt{\frac{2}{2\sqrt{c^3}}}=$

$=\big(\frac{1}{a}\big)^{\frac{3}{4}}+\big(\frac{1}{b}\big)^{\frac{3}{4}}+\big(\frac{1}{c}\big)^{\frac{3}{4}} \le 3\cdot \big(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \big)^{\frac{3}{4}}=3$

$$\frac{1}{\sqrt{a^3+1}} \leq \frac{1}{\sqrt{2a\sqrt{a}}}=\frac{\sqrt[4]{a^3}}{\sqrt{2}a}\leq \frac{a+a+a+1}{4\sqrt{2}a}=\frac{3a+1}{4\sqrt{2}a}=\frac{3}{4\sqrt{2}}+ \frac{1}{4\sqrt{2}}\cdot \frac{1}{a}$$

$${1\over \sqrt{a^3+1}}+{1\over \sqrt{b^3+1}}+{1\over \sqrt{c^3+1}} \leq \frac{3}{4\sqrt{2}}+ \frac{3}{4\sqrt{2}}+ \frac{3}{4\sqrt{2}}+ \frac{1}{4\sqrt{2}}\cdot \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big)={3\over \sqrt{2}}$$

$$a\leq b \leq c \quad \Longleftrightarrow \quad \frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c} \quad \Longleftrightarrow \quad 3=\frac{1}{a}+ \frac{1}{b} + \frac{1}{c} \geq \frac{1}{a} +\frac{1}{a} +\frac{1}{a} =\frac{3}{a} \quad \Rightarrow c\geq b\geq a \geq1$$

$$\forall a\geq 1: \quad a^3+1\geq a^2+a \quad \Longleftrightarrow \quad \frac{1}{\sqrt{a^3+1}}\leq \frac{1}{\sqrt{a^2+a}}\leq \frac{1}{\sqrt{2}}$$

$$\Rightarrow \quad {1\over \sqrt{a^3+1}}+{1\over \sqrt{b^3+1}}+{1\over \sqrt{c^3+1}} \leq {1\over \sqrt{2}}+{1\over \sqrt{2}}+{1\over \sqrt{2}}={3\over \sqrt{2}} .$$

Қате мына жерде: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{1}{a}+\frac{1}{a}+\frac{1}{a}$