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[不等式] 二次根式函数值域

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力工 Posted 2018-6-5 07:19 |Read mode
证明:函数$ f(x,y)=\sqrt{x-y+1}+\sqrt{2x+y-2}+\sqrt{2-x}$的值域为$[\sqrt{\dfrac{5}{3}},\sqrt{\dfrac{35}{3}}]$.

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kuing Posted 2018-6-5 15:11
无难度,由 CS 有
\begin{align*}
f(x,y)&\leqslant \sqrt {(1+1)(x-y+1+2x+y-2)}+\sqrt {2-x}\\
&=\sqrt {6x-2}+\sqrt {2-x}\\
&\leqslant \sqrt {(6+1)\left( {x-\frac 13+2-x} \right)}\\
&=\sqrt {\frac {35}3},
\end{align*}
由 `\sqrt a+\sqrt b\geqslant\sqrt{a+b}` 有
\begin{align*}
f(x,y)&\geqslant \sqrt {x-y+1+2x+y-2}+\sqrt {2-x}\\
&=\sqrt {3x-1}+\sqrt {2-x}\\
&\geqslant \sqrt {x-\frac 13}+\sqrt {2-x}\\
&\geqslant \sqrt {x-\frac 13+2-x}\\
&=\sqrt {\frac 53},
\end{align*}
取等略。

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isee Posted 2018-6-5 15:51
回复 2# kuing

CS 原来是 柯西啊

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kuing Posted 2018-6-5 15:55
回复 3# isee

我在撸题集不是一开始就注明过的么……

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敬畏数学 Posted 2018-6-5 20:10
此题求最小值时,x=
1/3取最小值,再放缩。有难度。

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