Lambert W 上界

hbghlyj

对于每个 $ y>1/e $ 和 $ x\geq -1/e $，$$ W_{0}(x)\leq \ln \left({\frac {x+y}{1+\ln(y)}}\right), $$
仅对于 $ x=y\ln(y) $ 相等。

Czhang271828

原不等式等价于 $(1+\ln y)e^{W_0(x)}\leq x+y=W_0(x)e^{W_0(x)}+y$​, 即
$$
e^{W_0(x)}(W_0(x)-1-\ln y) +y\geq 0.
$$
记 $u:=W_0(x)$, 则
$$
\begin{align*}
e^u(u-1-\ln y)+y&\geq e^{u-\ln y}\cdot y(u-1-\ln y)+y\\
&\geq y(1+u-\ln y)(u-1-\ln y)+y\\
&=y(u-\ln y)^2.
\end{align*}
$$
取等条件 $u=W_0(x)=\ln y$, 即, $x=y\ln y$.

Czhang271828

Czhang271828 发表于 2023-5-29 21:29
原不等式等价于 $(1+\ln y)e^{W_0(x)}\leq x+y=W_0(x)e^{W_0(x)}+y$​, 即
$$
e^{W_0(x)}(W_0(x)-1-\ln y)  ...

置 $y=x^t/e$, 则原不等式化作
$$
W_0(x)\leq \ln \dfrac{x+x^t/e}{t\ln x}=\ln x-\ln\ln x-\ln t+\ln(1+x^{t-1}/e).
$$
例如 $t=1$ 时有 $\boxed{W_0(x)\leq \ln x-\ln\ln x+\ln (1+e^{-1})}$. 最小值在 $t\ln x=1$​ 处取达, 回代就是恒等式.

hbghlyj

Bounds and inequalities
Hoorfar and Hassani[10] showed that the following bound holds for $x ≥ e$:$$ \ln x-\ln \ln x+{\frac {\ln \ln x}{2\ln x}}\leq W_{0}(x)\leq \ln x-\ln \ln x+{\frac {e}{e-1}}{\frac {\ln \ln x}{\ln x}}. $$
Roberto Iacono and John P. Boyd[12] enhanced the bounds as following:
$$ \ln \left({\frac {x}{\ln x}}\right)-{\frac {\ln \left({\frac {x}{\ln x}}\right)}{1+\ln \left({\frac {x}{\ln x}}\right)}}\ln \left(1-{\frac {\ln \ln x}{\ln x}}\right)\leq W_{0}(x)\leq \ln \left({\frac {x}{\ln x}}\right)-\ln \left(\left(1-{\frac {\ln \ln x}{\ln x}}\right)\left(1-{\frac {\ln \left(1-{\frac {\ln \ln x}{\ln x}}\right)}{1+\ln \left({\frac {x}{\ln x}}\right)}}\right)\right). $$