一元平均值的不等式

ljh25252

\begin{align*}
&\color{gray}{A=\frac{x+1}{2}\quad I=\frac{1}{e}\cdot x^{\frac{x}{x-1}}\quad L=\frac{x-1}{\ln x}\quad G=\sqrt{x}\quad H=\frac{2x}{x+1}}\\
\end{align*}
\[\color{red}{\frac{eI-2A}{e-2}\geq \frac{G^2}{L}}\]
首发,不知道能不能显示,以防补上一个:
设f(x)=x^(x/(x-1))
求证:f(x)-x-1>=(e-2)lnf(x).

ljh25252

有另一边
\[\color{red}{8\cdot\dfrac{x-x^{\frac{8}{9}}}{x^\frac{8}{9}-1}>\frac{eI-2A}{e-2}>\frac{G^2}{L}.}\]

ljh25252

把最近群里讨论的一元均值都发了,看看有谁做一下

ljh25252

$$
\color{grey}{A=\frac{x+1}{2}\quad G=\sqrt{x}\quad H=\frac{2x}{x+1} \quad L=\frac{x-1}{\ln x}\quad I=\frac{1}{e}\cdot x^{\frac{x}{x-1}}\quad}\\ \color{grey}{M_p=\left(\frac{x^p+1}{2}\right)^{\frac{1}{p}}}
$$
$$
\begin{align} 1.&\frac{(1-x)^2}{3(1+x)^2}\leq\ln\frac{I}{G}\leq\frac{(1-x)^2}{12x}\\ 2.&\frac{(1-x)^2}{2(1+x)^2}\leq\ln\frac{A}{G}\leq\frac{(1-x)^2}{8x}\\ 3.&AG\leq\frac{G}{2L}(A^2+L^2)\leq IL\\ 4.&\ln\left(\frac{A_{\sqrt{x}}^4}{I^2_{\sqrt{x}}\cdot I_{x}}\right)\leq\frac{2G_x+A_x}{L_x}\\ 5.&A^2_x\leq I_{x^2}\leq \overline{G}^2_x\quad\color{gray}{(\overline{G}=x^{\frac{x}{1+x}})}\\ 6.&I^3_{x^3}\leq\overline{G}^2_x\cdot A^4_x\\ 7.&I^2_{x^2}\leq\frac{A^4_x}{I_x^2}\leq\frac{I^4_x}{G^2_x}\\ 8.&L^2\leq I\cdot\left(\frac{A+G}{2}\right)\cdot\exp\left(\frac{G-L}{L}\right)\\ 9.&L\leq I\cdot\exp\left(\frac{G-L}{L}\right)\\ \end{align}\\
$$

ljh25252

$$
\color{#A00}{2A+H\geq3L\Leftrightarrow 2\frac{x+1}{2}+\frac{2x}{x+1}\geq3\frac{x-1}{\ln x}}\\
$$

ljh25252

$$
\color{#F0F}{(eI-2A)L\geq (e-2)G^2\Leftrightarrow (x^\frac{x}{x-1}-x-1)\cdot\frac{x-1}{\ln x}\geq (e-2)x}\\
$$

ljh25252

$$
\color{#F0A}{2A+G\leq I\Leftrightarrow x+\sqrt{x}+1\leq\frac{1}{e}x^{\frac{x}{x-1}}}\\
$$

ljh25252

$$
\color{#A0F}{L+I\leq A+G\Leftrightarrow \frac{x-1}{\ln x}+\frac{1}{e}x^\frac{x}{x-1}\leq \frac{x+1}{2}\cdot \sqrt{x}}\\
$$

ljh25252

$$
\color{#A08}{AG\leq LI\Leftrightarrow \frac{x+1}{2}\cdot\sqrt{x}\leq \frac{x-1}{\ln x}\cdot\frac{1}{e}\cdot x^{\frac{x}{x-1}}}\\
$$

ljh25252

$$
\color{blue}{L\leq H^L\leq e^{L-1}}
$$
$$
\color{grey}{ A=\frac{x+1}{2}\quad H=\frac{2x}{x+1}\quad  L=\frac{x-1}{\ln x}}\\
$$

ljh25252

$$
\color{blue}{\frac{\ln G}{\ln H}\leq \frac{L+1}{2}}
$$
$$
\color{grey}{A=\frac{x+1}{2}\quad L=\frac{x-1}{\ln x}\quad G=\sqrt{x}\quad H=\frac{2x}{x+1}}\\
$$

ljh25252

$$
1+L\ln L<2A-L<2(1+L\ln L)-A
$$
$$
\color{grey}{A=\frac{x+1}{2}\quad L=\frac{x-1}{\ln x}}\\
$$

ljh25252

$$
Le\geq e^{\frac{1}{L}}(L-A+1)(L+A-1)
$$
$$
\color{grey}{A=\frac{x+1}{2}\quad L=\frac{x-1}{\ln x}}\\
$$

Reinhardt

好强

ljh25252

$$
\begin{align}
x\in(0,1)\quad &L(L(x))\leq A(G(x))\\
x\in(0,1)\quad &L^2\leq AG\\
\end{align}
$$
$$
\color{gray}{A=\frac{x+1}{2}\quad L=\frac{x-1}{\ln x}\quad G=\sqrt{x}}
$$

ljh25252

$$
ex^\frac{1}{1-x}\cdot\frac{1-x}{1+x}<\dfrac{x-1-\ln x}{x+\frac{1}{x}-2}<\dfrac{x^{\frac{2}{3}}}{1+x^{\frac{2}{3}}}
$$

ljh25252

$$
0<x<a\quad e^x\ln x<\frac{x^a}{a-x}
$$

ljh25252

$$
ex^{\frac{1}{1-x}}\left(\frac{1-x}{1+x}\right)^2<(x-1-\ln x)(1+x)\ln(1+x)\quad x\in(0,1)
$$

ljh25252

$$
(1-k)G^k+kA^k\geq A_{x^k}\quad k\in(0,1)
$$

ljh25252

$$
\sqrt{1+\dfrac{1}{x+\dfrac{1}{e^2-2}}}<\dfrac{e}{\left(1+\frac{1}{x}\right)^x}<\sqrt{1+\dfrac{1}{x+\frac{1}{e^2-1}}}\quad (x>0)
$$