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[不等式] 求最小值

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v6mm131 Posted 2015-2-6 11:32 |Read mode
在面积为1的所有锐角三角形中,找到一个三角形使得表达式$P=\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}$取得最小值。
oh 为了说清楚 还是指明一下$a,b,c$是三角形的三边

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kuing Posted 2015-2-6 14:28
分明就是伊朗96

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kuing Posted 2015-2-6 15:00
由锐角三角形可令 $a^2=y+z$, $b^2=z+x$, $c^2=x+y$, $x$, $y$, $z>0$,代入海伦-秦九韶公式中得 $xy+yz+zx=4$,而 $P=\sum\frac1{(y+z)^2}$,于是就是伊朗96了

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 Author| v6mm131 Posted 2015-2-6 16:17
明白了

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其妙 Posted 2015-2-6 16:53
由锐角三角形可令 $a^2=y+z$, $b^2=z+x$, $c^2=x+y$, $x$, $y$, $z>0$,代入海伦-秦九韶公式中得 $xy+yz+z ...
kuing 发表于 2015-2-6 15:00
牛神笔!这都看的出?

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kuing Posted 2015-2-6 17:18
回复 5# 其妙

?很自然啊,只要想到用那个换元把锐角三角形化掉,之后就水到渠成了

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