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Last edited by ljh25252 at 2024-12-25 19:53:00对于
$$
\begin{aligned} &\frac{1}{a^{2k}}+\frac{1}{b^{2k}}+\frac{1}{c^{2k}}\geq\frac{2^{1-k}+\frac{1}{2^{2k}}}{S^k}\\ \Leftrightarrow &\sum\frac{S^k}{a^{2k}}=\sum\left(\frac{S}{a^2}\right)^k=\sum\left(\frac{\frac{1}{2}ab\sin C}{a}\right)^k=\frac{1}{2^k}\sum\left(\frac{\sin B\sin C}{\sin A}\right)^k\geq\frac{2}{2^k}+\left(\frac{1}{2^k}\right)^2\\ \Leftrightarrow&\sum\left(\frac{\sin B\sin C}{\sin A}\right)^k\geq2+\frac{1}{2^k} \end{aligned}
$$
置 $(x,y,z)\to(\cot A,\cot B,\cot C)$ ，有 $xy+yz+zx=1$ ，化为
$$
\sum\frac{1}{(x+y)^k}\geq2+\frac{1}{2^k}
$$
当 $k=\frac{1}{2}$ 时，有
$$
\sum\frac{1}{\sqrt{x+y}}\geq2+\frac{1}{\sqrt{2}}
$$
不妨设 $x\geq y\geq z$ 并置 $x=\frac{1-yz}{y+z}$ 有 $y+z\leq 2,yz\leq\frac{1}{3}$ 以及恒等式如下：
$$
\frac{1}{y^2+1}+\frac{1}{z^2+1}-\frac{1}{(y+z)^2+1}-1=\frac{yz[2-2yz-yz(y+z)^2]}{(y^2+1)(z^2+1)[(y+z)^2+1]}\geq0\\ \Rightarrow \frac{1}{1+y^2}+\frac{1}{1+z^2}\geq 1+\frac{1}{1+(y+z)^2}
$$
$$
1+(y+z)^2-(1+y^2)(1+z^2)=yz(2-yz)\\ \Rightarrow 1+(y+z)^2\geq (1+y^2)(1+z^2)
$$
对原式使用即：
$$
\begin{aligned}
\mathrm{L.H.S}&=\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}\\
&=\frac{1}{\sqrt{y+z}}+\sqrt{y+z}\left(\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\right)\\
&=\frac{1}{\sqrt{y+z}}+\sqrt{y+z}\sqrt{\left(\frac{1}{1+y^2}+\frac{1}{1+z^2}+\frac{2}{\sqrt{(1+y^2)(1+z^2)}}\right)}\\
&\geq\frac{1}{\sqrt{y+z}}+\sqrt{y+z}\sqrt{\left(1+\frac{1}{1+(y+z)^2}+\frac{2}{\sqrt{1+(y+z)^2}}\right)}\\
&=\frac{1}{\sqrt{y+z}}+\sqrt{y+z}\left(1+\frac{1}{\sqrt{1+(y+z)^2}}\right)\\
&=\frac{1}{\sqrt{t}}+\sqrt{t}+\frac{1}{\sqrt{t+\frac{1}{t}}}\qquad \color{gray}{(t:=y+z)}\\
&\geq2+\frac{1}{\sqrt{2}}
\end{aligned}
$$
当 $k=1$ 时，有
$$
\sum\frac{1}{x+y}-\frac{5}{2}=\frac{4(x+y+z-2)^2+3x(y+z-1)^2+3y(z+x-1)^2+3z(x+y-1)^2+xyz}{4(x+y)(y+z)(z+x)}\geq0
$$
当 $k=2$ 时，有
$$
(xy+yz+zx)\sum\frac{1}{(x+y)^2}-\frac{9}{4}=\frac{2xyz(x^2+y^2+z^2+3xyz-x^2(y+z)-y^2(z+x)-z^2(x+y))+\sum xy(4(x+y)^2-xy)(x-y)^2}{4(x+y)^2(y+z)^2(z+x)^2}\geq0
$$
对于 $k=3$ ，取 $(x,y,z)\to\left(\frac{1}{\sqrt{5}},\frac{3}{2\sqrt{5}},\frac{7}{5\sqrt{5}}\right)$ 有 $xy+yz+zx=1$ 且
$$
\sum\frac{1}{(x+y)^3}=\frac{934231661}{210720960\sqrt{5}}=\sqrt{\frac{872788796414818921}{222016614916608000}}\leq\sqrt\frac{888066459666432000}{222016614916608000}=\sqrt{4}\leq2+\frac{1}{8}
$$
原不等式不成立。 |
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ljh25252
Posted at 2024-12-23 19:31:44
kuing
Posted at 2024-12-25 14:46:42
