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[不等式] 用柯西怎么凑系数?

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facebooker Posted 2022-9-18 02:51 |Read mode
求$\sqrt{2+x}+\sqrt{14+3x}+2\sqrt{4-2x}$的取值范围?

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kuing Posted 2022-9-18 03:25
`\sqrt{kx+b}` 一定是上凸函数,上凸函数之和仍为上凸函数,所以最小值一定在定义域端点,比较一下两端大小即可,我就不算了。

最大值柯西可以待定系数
\begin{align*}
&=\sqrt{2+x}+\sqrt{\frac1a(14a+3ax)}+\sqrt{\frac1b(16b-8bx)}\\
&\leqslant\sqrt{\left(1+\frac1a+\frac1b\right)(2+x+14a+3ax+16b-8bx)},
\end{align*}
由取等条件及 `x` 系数为零组成方程组
\[\led
&2+x=a^2(14+3x)=b^2(16-8x),\\
&1+3a-8b=0,
\endled\]
只要不是钓鱼题或者抄错题,这个方程组就应该有简单解,时间关系,我也不算了。

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凌晨3点半  Posted 2022-9-18 18:15

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isee Posted 2022-9-18 18:46
根式下一次,求最大值,凑柯西不等式

\begin{align*}
f(x)&=\sqrt {2+x}+\sqrt{3\left(\frac {14}3+x\right)}+\sqrt{4(4-2x)}\\[1.5em]
&\quad\,(1+3+4)\left(2+x+\frac {14}3+x+4-2x\right)\\[1ex]
&\geqslant\left(\sqrt {2+x}+\sqrt{14+3x}+\sqrt{16-8x}\right)^2\\[1ex]
\Rightarrow f(x)&\leqslant \frac {16\sqrt 3}{3}.
\end{align*}
取 “=” 时,$x=-2/3$.
isee=freeMaths@知乎

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