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[不等式] 二次函数不等式

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hjfmhh posted 2019-3-5 10:14 |Read mode
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kuing posted 2019-3-5 15:53
只是看起来吓人,其实没什么难度。

引理:设 `g(m)=\sqrt{m+x}\sqrt{m+y}`,则 `g'(m)\geqslant1`。(求导即可证,过程不码了)

回到原题,依题意可设 `f(x)=p(x-k)^2+m`,且 `2m\geqslant f(1)=p(1-k)^2+m`,即 `m\geqslant p(1-k)^2`,原不等式等价于
\[\sum\sqrt{p(a-k)^2+m}\sqrt{p(b-k)^2+m}\geqslant3p(1-k)^2+3m,\]若令 `G(m)=` 上式的左边减右边,则由引理可知 `G'(m)\geqslant0`,从而只需证明当 `m=p(1-k)^2` 的情形即可,此时不等式化为
\[\sum\sqrt{(a-k)^2+(1-k)^2}\sqrt{(b-k)^2+(1-k)^2}\geqslant6(1-k)^2,\]由柯西有
\[\sqrt{(a-k)^2+(1-k)^2}\sqrt{(1-k)^2+(b-k)^2}\geqslant(\abs{a-k}+\abs{b-k})\abs{1-k},\]所以
\[\LHS\geqslant2(\abs{a-k}+\abs{b-k}+\abs{c-k})\abs{1-k}\geqslant2\abs{(a+b+c-3k)(1-k)}=\RHS.\]

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original poster hjfmhh posted 2019-3-5 21:27
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kuing启发下,直接cs也行

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