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[不等式] 求证一个三元不等式

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lemondian Posted 2019-8-27 00:02 |Read mode
设$a,b,c$均为正实数,求证:
$\dfrac{a(a^2+bc)}{b+c}+\dfrac{b(b^2+ca)}{c+a}+\dfrac{c(c^2+ab)}{a+b}\geqslant ab+bc+ca$.

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kuing Posted 2019-8-27 00:39
感觉以前撸过,不过懒得找了,反正简单
\begin{align*}
&\iff\sum\left( \frac{a(a^2+bc)}{b+c}+a^2+bc \right)\geqslant\sum a^2+2\sum ab\\
&\iff\sum\frac{(a+b+c)(a^2+bc)}{b+c}\geqslant(a+b+c)^2\\
&\iff\sum\frac{a^2+bc}{b+c}\geqslant a+b+c\\
&\iff\sum\left( \frac{a^2+bc}{b+c}+a \right)\geqslant2(a+b+c)\\
&\iff\sum\frac{(a+b)(a+c)}{b+c}\geqslant2(a+b+c)\\
&\liff\frac{yz}x+\frac{zx}y+\frac{xy}z\geqslant x+y+z
\end{align*}

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其妙 Posted 2019-8-27 16:27
和下面这个不等式有何关系?能串联起来吗?
blog6.jpg
妙不可言,不明其妙,不着一字,各释其妙!

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 Author| lemondian Posted 2019-8-27 22:15
回复 2# kuing
唉,最后一行我都不会证呀

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