2022年江西预赛不等式解答

(2022年江西预赛)设$x,y,z$为正实数,满足$xyz=1$.证明: $\sqrt{1+8x}+\sqrt{1+8y}+\sqrt{1+8z}\geqslant 9$.

证法一.由均值不等式可得
$$\begin{aligned} &\sqrt{1+8x}+\sqrt{1+8y}+\sqrt{1+8z} \geqslant 3\sqrt[3]{\sqrt{1+8x}\cdot \sqrt{1+8y}\cdot \sqrt{1+8z}} \\ &=3\sqrt[6]{1+8\left( x+y+z \right) +64\left( xy+yz+zx \right) +8^3xyz} \\ &\geqslant 3\sqrt[6]{1+8\cdot 3\sqrt[3]{xyz}+64\cdot 3\sqrt[3]{x^2y^2z^2}+8^3xyz} \\ &=3\sqrt[6]{1+8\cdot 3+64\cdot 3+8^3}=3\sqrt[6]{\left( 1+8 \right) ^3}=9. \end{aligned}$$

证法二. (熊孩子数学公众号)注意到
$$\sqrt{1+8x}=\sqrt{1+x+x+\cdots +x}\geqslant \sqrt{9\sqrt[9]{x^8}}=3\sqrt[18]{x^8}=3\sqrt[9]{x^4},$$
则
$$\begin{aligned} \sqrt{1+8x}+\sqrt{1+8y}+\sqrt{1+8z} &\geqslant 3\left( \sqrt[9]{x^4}+\sqrt[9]{y^4}+\sqrt[9]{z^4} \right) \\ &\geqslant 9\sqrt[3]{\sqrt[9]{x^4}\cdot \sqrt[9]{y^4}\cdot \sqrt[9]{z^4}}=9\sqrt[27]{x^4y^4z^4}=9. \end{aligned}$$

证法三. (熊孩子数学公众号)利用待定系数找到$k,b$使得
$$f(x)=\sqrt{1+8x}\geqslant k\ln x+b=g(x).$$
注意到不等式的取等条件为$x=y=z=1$,故只需保证$f'(1)=g'(1)$和$f(1)=g(1)$,由此得$\displaystyle k=\frac{4}{3},b=3$.

下面证明:
$$\sqrt{1+8x}\geqslant \frac{4}{3}\ln x+3.$$

令$\displaystyle h(x)=\sqrt{1+8x}-\frac{4}{3}\ln x-3$,则
$$\begin{aligned} h'\left( x \right) &=\frac{4}{\sqrt{1+8x}}-\frac{4}{3x}=\frac{4\left( 3x-\sqrt{1+8x} \right)}{3x\sqrt{1+8x}} \\ &=\frac{4\left( 9x^2-8x-1 \right)}{3x\sqrt{1+8x}\left( 3x-\sqrt{1+8x} \right)}=\frac{4\left( x-1 \right) \left( 9x+1 \right)}{3x\sqrt{1+8x}\left( 3x-\sqrt{1+8x} \right)}. \end{aligned}$$
故$h(x)$在$(0,1)$上递减,在$[1,+\infty)$上递增.
则$h(x)\geqslant h(1)=0$.即$\displaystyle\sqrt{1+8x}\geqslant \frac{4}{3}\ln x+3$.

于是
$$\sqrt{1+8x}+\sqrt{1+8y}+\sqrt{1+8z}\geqslant \frac{4}{3}\ln \left( xyz \right) +9=9.$$

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(朋友的端午节礼物)已知函数$\displaystyle f\left( x \right) =a\sqrt{x-\frac{1}{4}}+\frac{b}{2}-e^{\sqrt{x}}$有零点,则当$a^2+b^2$取最小值时, $\displaystyle\frac{b}{a}=$?
解.函数的定义域为$\displaystyle\left[ \frac{1}{4},+\infty \right)$.

 

(1)当$a\leqslant 0$时, $f(x)$单调递减,只需$\displaystyle f\left( \frac{1}{4} \right) =\frac{b}{2}-\sqrt{e}\geqslant 0$,即$b\geqslant 2\sqrt{e}$.此时$a^2+b^2\geqslant 4e$.

(2)当$a>0$时,
$$f'\left( x \right) =\frac{a-\sqrt{1-\frac{1}{4x}}e^{\sqrt{x}}}{2\sqrt{x-\frac{1}{4}}},$$
注意到$\displaystyle g(x)=a-\sqrt{1-\frac{1}{4x}}e^{\sqrt{x}}$在$\displaystyle\left[ \frac{1}{4},+\infty \right)$上单调递减,而$\displaystyle g\left( \frac{1}{4} \right) =a>0$,当$x\to +\infty$时, $g(x)\to-\infty$.则$g(x)$在$\displaystyle\left[ \frac{1}{4},+\infty \right)$上存在唯一零点$x_0$,此为$f(x)$的极大值点,满足
$$a=\sqrt{1-\frac{1}{4x_0}}e^{\sqrt{x_0}}.$$

故只需
$$f\left( x_0 \right) =a\sqrt{x_0-\frac{1}{4}}+\frac{b}{2}-e^{\sqrt{x_0}}=\left( \sqrt{x_0}-\frac{1}{4\sqrt{x_0}}-1 \right) e^{\sqrt{x_0}}+\frac{b}{2}\geqslant 0,$$
则
$$b\geqslant \left( 2+\frac{1}{2\sqrt{x_0}}-2\sqrt{x_0} \right) e^{\sqrt{x_0}}.$$

(a)若$\displaystyle\frac{1}{4}\leqslant x_0\leqslant \frac{3+2\sqrt{2}}{4}$时, $\displaystyle 2+\frac{1}{2\sqrt{x_0}}-2\sqrt{x_0}\geqslant 0$,则
$$\begin{aligned} a^2+b^2 &\geqslant \left( \sqrt{1-\frac{1}{4x_0}}e^{\sqrt{x_0}} \right) ^2+\left[ \left( 2+\frac{1}{2\sqrt{x_0}}-2\sqrt{x_0} \right) e^{\sqrt{x_0}} \right] ^2 \\ &=\left( 3+4x_0+\frac{2}{\sqrt{x_0}}-8\sqrt{x_0} \right) e^{2\sqrt{x_0}}, \end{aligned}$$
令
$$h\left( x \right) =\left( 3+4x+\frac{2}{\sqrt{x}}-8\sqrt{x} \right) e^{2\sqrt{x}},$$
则
$$h'\left( x \right) =\frac{\left( 4x^2-4x^{3/2}-x+2\sqrt{x}-1 \right) e^{2\sqrt{x}}}{x^{3/2}}=\frac{\left( \sqrt{x}-1 \right) \left( 4x-1+\frac{1}{\sqrt{x}} \right) e^{2\sqrt{x}}}{x},$$
而
$$4x-1+\frac{1}{\sqrt{x}}=4x+\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}-1\geqslant 3\sqrt[3]{4x\cdot \frac{1}{2\sqrt{x}}\cdot \frac{1}{2\sqrt{x}}}-1=2>0,$$
故$h(x)$在$\displaystyle\left[ \frac{1}{4},\frac{3+2\sqrt{2}}{4} \right]$上有极小值$h(1)=e^2$,此时$\displaystyle x_0=1,a=\sqrt{1-\frac{1}{4x_0}}e^{\sqrt{x_0}}=\frac{\sqrt{3}}{2}e,b=\left( 2+\frac{1}{2\sqrt{x_0}}-2\sqrt{x_0} \right) e^{\sqrt{x_0}}=\frac{1}{2}e$,则$\displaystyle\frac{b}{a}=\frac{\sqrt{3}}{3}$.

(b)若$\displaystyle x_0>\frac{3+2\sqrt{2}}{4}$时, $\displaystyle 2+\frac{1}{2\sqrt{x_0}}-2\sqrt{x_0}<0$,则
$$\begin{aligned} a^2+b^2 &\geqslant \left( \sqrt{1-\frac{1}{4x_0}}e^{\sqrt{x_0}} \right) ^2+0^2=\left( 1-\frac{1}{4x_0} \right) e^{2\sqrt{x_0}} \\ &>\left( 1-\frac{1}{4\cdot \frac{3+2\sqrt{2}}{4}} \right) e^{2\sqrt{\frac{3+2\sqrt{2}}{4}}}=\left( 2\sqrt{2}-2 \right) e^{\sqrt{2}+1}>e^2. \end{aligned}$$
此时$a^2+b^2$取不到最小值.

综上所述,当$\displaystyle x_0=1,a=\frac{\sqrt{3}}{2}e,b=\frac{1}{2}e$时$a^2+b^2$取到最小值,则$\displaystyle\frac{b}{a}=\frac{\sqrt{3}}{3}$.