Математикадан аудандық олимпиада, 2019-2020 оқу жылы, 9 сынып

$abc=2$ шартын қанағаттандыратын кез келген оң нақты $a,b,c$ сандары үшін $a^3+b^3+c^3 \ge a \sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$ теңсіздігін дәлелдеңіз.

Комментарий/решение

$a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b} \leqslant \sqrt{(a^2+b^2+c^2)((\sqrt{b+c})^2+(\sqrt{c+a})^2+(\sqrt{a+b})^2)}=\sqrt{2(a^2+b^2+c^2)(a+b+c)}=\sqrt{abc(a^2+b^2+c^2)(a+b+c)}$

$(a^3+b^3+c^3)^2 \geqslant abc(a^2+b^2+c^2)(a+b+c)$

$\dfrac{1}{2}T_{(6,\,0,\,0)}(x,\,y,\,z)+T_{(3,\,3,\,0)}(x,\,y,\,z) \geqslant \dfrac{1}{2}T_{(4,\,1,\,1)}(x,\,y,\,z)+T_{(3,\,2,\,1)}(x,\,y,\,z)$

$$\forall a,b,c >0,\quad abc=2 \qquad \qquad a^3+b^3+c^3 \geq a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b} $$

$\textbf{Решение:}$ $$2a^3+2b^3+2c^3 = a^3+b^3+c^3+\frac{1}{2}\Bigg(\frac{a^3+a^3+b^3}{3}+\frac{a^3+a^3+c^3}{3}+\frac{b^3+b^3+a^3}{3}+ \frac{b^3+b^3+c^3}{3}+\frac{c^3+c^3+a^3}{3}+\frac{c^3+c^3+b^3}{3}\Bigg)\geq$$

$$\geq a^3+b^3+c^3+\frac{a^2b+a^2c+b^2a+b^2c+c^2a+c^2b}{2} =a^3+b^3+c^3+\frac{ab(a+b)+bc(b+c)+ac(a+c)}{2}=$$

$$ =a^3+b^3+c^3+\frac{ab(a+b)+bc(b+c)+ac(a+c)}{abc}=a^3+b^3+c^3+\frac{a+b}{c}+\frac{b+c}{a} +\frac{a+c}{b}$$

$$= \Big( a^3+\frac{b+c}{a}\Big)+\Big( b^3+\frac{a+c}{b}\Big)+\Big( c^3+\frac{a+b}{c}\Big)\geq 2\sqrt{ a^3\cdot\frac{b+c}{a}}+2\sqrt{b^3\cdot\frac{a+c}{b}}+2\sqrt{c^3\cdot\frac{a+b}{c}}=$$

$$= 2a\sqrt{b+c}+2b\sqrt{a+c}+2c\sqrt{a+b}\Longleftrightarrow 2a^3+2b^3+2c^3 \geq 2a\sqrt{b+c}+2b\sqrt{a+c}+2c\sqrt{a+b}\Longleftrightarrow$$

$$\Longleftrightarrow a^3+b^3+c^3 \geq a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}$$

$2(a^3+b^3+c^3)\ge a^2(b+c)+b^2(c+a)+c^2(a+b)$ және $ \ \ a^3+b^3+c^3\ge 3abc=6 \ \ $ болғандықтан,

$$4(a^3+b^3+c^3)\ge 4+a^2(b+c)+4+b^2(c+a)+4+c^2(a+b)\ge $$

$$\ge 2\sqrt{4\cdot a^2(b+c)}+2\sqrt{4\cdot b^2(c+a)}+2\sqrt{4\cdot c^2(a+b)}=4\big( a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b} \big)$$

Лемма:

$a^3+b^3 \geq \frac{2(a+b)}{c}$

Дәлел:

$a^3+b^3=\left(\frac{a^3}{3}+\frac{a^3}{3}+\frac{b^3}{3}\right)+\left(\frac{b^3}{3}+\frac{b^3}{3}+\frac{a^3}{3}\right)\geq3\sqrt[3]{\frac{a^3}{3}\times\frac{a^3}{3}\times\frac{b^3}{3}}+3\sqrt[3]{\frac{b^3}{3}\times\frac{b^3}{3}\times\frac{a^3}{3}} = a^2b+b^2a=ab(a+b)=\frac{2}{c}(a+b)=\frac{2(a+b)}{c}$

Дәлелденді.

Жоғарыдағы лемма бойынша:

$a^3+b^3+c^3=\frac{1}{4}((a^3+b^3)+(b^3+c^3)+(c^3+a^3))+\left(\frac{a^3}{2}+\frac{b^3}{2}+\frac{c^3}{2}\right)\geq\frac{1}{4}\left(\frac{2(a+b)}{c}+\frac{2(b+c)}{a}+\frac{2(c+a)}{b}\right)+\left(\frac{a^3}{2}+\frac{b^3}{2}+\frac{c^3}{2}\right)=\left(\frac{b+c}{2a}+\frac{a^3}{2}\right)+\left(\frac{c+a}{2b}+\frac{b^3}{2}\right)+\left(\frac{a+b}{2c}+\frac{c^3}{2}\right)\geq2\sqrt{\frac{b+c}{2a}\times\frac{a^3}{2}}+2\sqrt{\frac{c+a}{2b}\times\frac{b^3}{2}}+2\sqrt{\frac{a+b}{2c}\times\frac{c^3}{2}}=a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$

$$\Leftrightarrow$$

$$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$$

As was to be shown.