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[不等式] 来自人教群:双三元不等式

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kuing posted 2024-12-26 23:51 |Read mode
渝A-教师-石崇的BOSS
0453DE60A10029CD8222830EEB749010.png
大神们,来活了
证明:由柯西有
\begin{align*}
\sum\left(\frac b{c+a}+\frac c{a+b}\right)x^2&=\sum\frac a{b+c}(y^2+z^2)\\
&=\sum\left(\frac{a+b+c}{b+c}-1\right)(y^2+z^2)\\
&=\frac12(b+c+c+a+a+b)\sum\frac{y^2+z^2}{b+c}-2\sum x^2\\
&\geqslant\frac12\left(\sum\sqrt{y^2+z^2}\right)^2-2\sum x^2\\
&=\sum\sqrt{y^2+z^2}\sqrt{x^2+z^2}-\sum x^2\\
&\geqslant\sum(xy+z^2)-\sum x^2\\
&=xy+yz+zx.
\end{align*}

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original poster kuing posted 2024-12-27 00:22
上面的证明实际上得到了加强式:`a`, `b`, `c`, `x`, `y`, `z>0` 则
\[\sum\frac a{b+c}(y+z)\geqslant\sum\sqrt{(z+x)(x+y)}-\sum x.\]
这是老题,我在《撸题集》P.480 里就提到过,不过当时我只说“有点眼熟”而没写证明,可能没想出来,其实 1# 的柯西证法也不算难想啊……

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