|
|
本帖最后由 kuing 于 2024-7-7 21:59 编辑 自动脑补 `a_i` 全为正。
和式太复杂鸟，先看看 `n=3` 时是怎么样的再说，此时等价于：`a`, `b`, `c`, `d>0`，求证
\begin{align*}
&\frac{\left(\sqrt{ab(\frac1a+\frac1b)}+\sqrt{ac(\frac1a+\frac1c)}+\sqrt{bc(\frac1b+\frac1c)}\right)^2}{abc(\frac1a+\frac1b+\frac1c)}+\frac{\left(\sqrt{ab(\frac1a+\frac1b)}+\sqrt{ad(\frac1a+\frac1d)}+\sqrt{bd(\frac1b+\frac1d)}\right)^2}{abd(\frac1a+\frac1b+\frac1d)}\\
&+\frac{\left(\sqrt{ac(\frac1a+\frac1c)}+\sqrt{ad(\frac1a+\frac1d)}+\sqrt{cd(\frac1c+\frac1d)}\right)^2}{acd(\frac1a+\frac1c+\frac1d)}+\frac{\left(\sqrt{bc(\frac1b+\frac1c)}+\sqrt{bd(\frac1b+\frac1d)}+\sqrt{cd(\frac1c+\frac1d)}\right)^2}{bcd(\frac1b+\frac1c+\frac1d)}\\
\leqslant{}&\frac{2\left(\sqrt{abc(\frac1a+\frac1b+\frac1c)}+\sqrt{abd(\frac1a+\frac1b+\frac1d)}+\sqrt{acd(\frac1a+\frac1c+\frac1d)}+\sqrt{bcd(\frac1b+\frac1c+\frac1d)}\right)^2}{abc+abd+acd+bcd},
\end{align*}
也就是
\begin{align*}
&\frac{\left(\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c}\right)^2}{ab+ac+bc}+\frac{\left(\sqrt{a+b}+\sqrt{a+d}+\sqrt{b+d}\right)^2}{ab+ad+bd}\\
&+\frac{\left(\sqrt{a+c}+\sqrt{a+d}+\sqrt{c+d}\right)^2}{ac+ad+cd}+\frac{\left(\sqrt{b+c}+\sqrt{b+d}+\sqrt{c+d}\right)^2}{bc+bd+cd}\\
\leqslant{}&\frac{2\left(\sqrt{ab+ac+bc}+\sqrt{ab+ad+bd}+\sqrt{ac+ad+cd}+\sqrt{bc+bd+cd}\right)^2}{abc+abd+acd+bcd}.
\end{align*}
如果作置换 `(a,b,c,d)\mapsto(1/a,1/b,1/c,1/d)`，则变成
\begin{align*}
&\frac{\left(\sqrt{c(a+b)}+\sqrt{b(a+c)}+\sqrt{a(b+c)}\right)^2}{a+b+c}+\frac{\left(\sqrt{d(a+b)}+\sqrt{b(a+d)}+\sqrt{a(b+d)}\right)^2}{a+b+d}\\
&+\frac{\left(\sqrt{d(a+c)}+\sqrt{c(a+d)}+\sqrt{a(c+d)}\right)^2}{a+c+d}+\frac{\left(\sqrt{d(b+c)}+\sqrt{c(b+d)}+\sqrt{b(c+d)}\right)^2}{b+c+d}\\
\leqslant{}&\frac{2\left(\sqrt{d(a+b+c)}+\sqrt{c(a+b+d)}+\sqrt{b(a+c+d)}+\sqrt{a(b+c+d)}\right)^2}{a+b+c+d}.
\end{align*}
作上述置换的意义在于：假如允许上式的变量取零，则取等条件有以下三种：
- `a=b=c=d>0`；
- `a=b=c>0` 且 `d=0`，及其轮换；
- 任意两个变量为零。
看来此题有难度，待续……🤔
|
|
kuing
发表于 2024-7-7 21:49
