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[不等式] 求不等式的代数解法

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chn-2000 Posted at 2023-8-14 08:07:26 |Read mode
求$\sqrt{x^2+y^2}$+$\sqrt{x^2-4x+4+y^2}$+$\sqrt{x^2+y^2-4y+4}$的最小值。

网上看见的解答都是采用数形结合换成为费马点模型,现求本题的代数解法。

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kuing Posted at 2023-8-14 10:25:22
令 `t=x+y`,则
\begin{align*}
\text{原式}&=\sqrt{x^2+y^2}+\sqrt{(x-2)^2+y^2}+\sqrt{(y-2)^2+x^2}\\
&\geqslant\frac{x+y}{\sqrt2}+\sqrt{(x-2+y-2)^2+(y+x)^2}\\
&=\frac t{\sqrt2}+\sqrt{(t-4)^2+t^2}\\
&=\frac t{\sqrt2}+\frac{\sqrt{(1+3)\bigl((2-t)^2+4\bigr)}}{\sqrt2}\\
&\geqslant\frac t{\sqrt2}+\frac{2-t+\sqrt{12}}{\sqrt2}\\
&=\sqrt2+\sqrt6,
\end{align*}
取等略。

PS、楼主代码的中间四个 \$ 可以省略。

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 Author| chn-2000 Posted at 2023-8-14 11:11:07
kuing 发表于 2023-8-14 10:25
令 `t=x+y`,则
\begin{align*}
\text{原式}&=\sqrt{x^2+y^2}+\sqrt{(x-2)^2+y^2}+\sqrt{(y-2)^2+x^2}\\
好的,感谢坛主!

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