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[不等式] 求最小值

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创之谜 Posted 2025-3-3 18:06 |Read mode
$已知x,y>0,x+3y=x^{3}y^{2}求\dfrac{3}{x}+\dfrac{2}{y}最小值?$

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isee Posted 2025-3-3 19:16
将条件改写为\[x^2=\frac3{xy}+\frac1{y^2},\]
于是由均值不等式,有
\begin{align*}
\left(\frac3x+\frac2y\right)^2&=\frac9{x^2}+4\left(\frac{3}{xy}+\frac1{y^2}\right)\\
&=\frac9{x^2}+4x^2\\
&\geqslant 12,
\end{align*}

再开方即知所求式的最小值为 $2\sqrt3$.
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kuing Posted 2025-3-3 19:23

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isee Posted 2025-3-3 19:25
kuing 发表于 2025-3-3 19:23
类似题:
https://kuing.cjhb.site/forum.php?mod=viewthread&tid=9031
https://kuing.cjhb.site/forum.php ...
原来这些年都过去了,已经~
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isee Posted 2025-3-4 17:24
kuing 发表于 2025-3-3 19:23
类似题:
……
在知乎得知此题来自《安徽省“江南十校”2025届高三下学期联考数学》第8题.
isee=freeMaths@知乎

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