（z）一个自然对数不等式

realnumber

江苏南京冯加明(12-----96)  18:36:40
本题除了加权琴生不等式，还可以怎么证？
设 $x, y, a, b$ 均为正数，求证：$x \ln \frac{x}{a}+y \ln \frac{y}{b} \geq(x+y) \ln \frac{x+y}{a+b}$

kuing

\begin{align*}
x\ln \frac xa+y\ln \frac yb\geqslant (x+y)\ln \frac{x+y}{a+b} &\iff \frac{x^xy^y}{a^xb^y}\geqslant \left( \frac{x+y}{a+b} \right)^{x+y} \\
&\iff \frac{(xy)^{x+y}}{(ay)^x(bx)^y}\geqslant \left( \frac{x+y}{a+b} \right)^{x+y} \\
&\iff \left( \frac{xy(a+b)}{x+y} \right)^{x+y}\geqslant (ay)^x(bx)^y \\
&\iff \frac x{x+y}ay+\frac y{x+y}bx\geqslant (ay)^{x/(x+y)}(bx)^{y/(x+y)},
\end{align*}
令 $x/(x+y)=\lambda\in(0,1)$, $ay=A$, $bx=B$，上式即
\[\lambda A+(1-\lambda)B\geqslant A^\lambda B^{1-\lambda} ,\]
也就是加权均值不等式了，导数神马的随便证……

kuing

加权均值能变成这么有型的对数不等式，也算是命题者的本事……

其妙

这不是young不等式吗？

realnumber

回复 4# 其妙
定理2（Young不等式）设 $p>1, q>1, \frac{1}{p}+\frac{1}{q}=1$ ，则 $\forall a, b \geq 0$ ，必有
$$
a \cdot b \leq \frac{a^p}{p}+\frac{b^q}{q}
$$
上式中等号成立的充要条件是 $a^p=b^q$ ．
换元一样

又，kk威武~~~~

kuing

名称没所谓，都是一样的东西

其妙

名称没所谓，都是一样的东西
kuing 发表于 2014-3-15 20:55

，又kk勇猛~~~~！

hbghlyj

realnumber 发表于 2014-3-15 12:50
定理2 (Young不等式) 设 $p>1, q>1, \frac{1}{p}+\frac{1}{q}=1$, 则 $\forall a, b \geq 0$, 必有
\[
a \cdot b \leq \frac{a^p}{p}+\frac{b^q}{q},
\]
上式中等号成立的充要条件是 $a^p=b^q$.

梅加强. 数学分析[M]. 2 版. 北京: 高等教育出版社, 2020.
例 3.5.13 (Young 不等式). 设 $f$ 是 $[0,+\infty)$ 中的单调递增连续函数, $f(0)=0$, $f^{-1}(y)$ 表示 $f$ 的反函数, 则当 $a, b>0$ 时
\[
a b \leqslant \int_0^a f(x) d x+\int_0^b f^{-1}(y) d y,
\]
等号成立当且仅当 $b=f(a)$.
证明. 我们分情况讨论.
(1) $b=f(a)$. 这时 $f:[0, a] \rightarrow[0, b]$ 为连续函数, 其逆 $f^{-1}:[0, b] \rightarrow[0, a]$ 也是连续函数. 取 $[0, a]$ 的 $n$ 等分:
\[
\pi: 0=x_0<x_1<\cdots<x_n=a,
\]
则 $y_i=f\left(x_i\right)(0 \leqslant i \leqslant n)$ 构成 $[0, b]$ 的分划:
\[
\pi^{\prime}: 0=f\left(x_0\right)=y_0<y_1<\cdots<y_n=b
\]
因为 $f$ 在闭区间上一致连续, 故当 $n \rightarrow \infty$ 时 $\left\|\pi^{\prime}\right\| \rightarrow 0$. 因此
\[
\begin{aligned}
& \int_0^a f(x) d x+\int_0^b f^{-1}(y) d y \\
& =\lim _{n \rightarrow \infty} \sum_{i=1}^n f\left(x_i\right)\left(x_i-x_{i-1}\right)+\lim _{n \rightarrow \infty} \sum_{i=1}^n f^{-1}\left(f\left(x_{i-1}\right)\right)\left(f\left(x_i\right)-f\left(x_{i-1}\right)\right) \\
& =\lim _{n \rightarrow \infty} \sum_{i=1}^n\left[f\left(x_i\right)\left(x_i-x_{i-1}\right)+x_{i-1}\left(f\left(x_i\right)-f\left(x_{i-1}\right)\right)\right] \\
& =\lim _{n \rightarrow \infty} \sum_{i=1}^n\left[f\left(x_i\right) x_i-x_{i-1} f\left(x_{i-1}\right)\right] \\
& =\lim _{n \rightarrow \infty}\left[x_n f\left(x_n\right)-x_0 f\left(x_0\right)\right] \\
& =a f(a)-0 f(0)=a f(a)=a b .
\end{aligned}
\]
(2) $0<b<f(a)$. 由 $f$ 的连续性, 存在 $\xi \in(0, a)$ 使得 $b=f(\xi)$. 由 (1), 有
\[
\begin{aligned}
& \int_0^a f(x) d x+\int_0^b f^{-1}(y) d y \\
& =\int_0^{\xi} f(x) d x+\int_{\xi}^a f(x) d x+\int_0^{f(\xi)} f^{-1}(y) d y \\
& >f(\xi)(a-\xi)+\int_0^{\xi} f(x) d x+\int_0^{f(\xi)} f^{-1}(y) d y \\
& =f(\xi)(a-\xi)+\xi f(\xi)=a f(\xi)=a b .
\end{aligned}
\]
(3) $b>f(a)$. 这时将 $f$ 视为 $f^{-1}$ 的反函数就可将问题化为情形 (2).

hbghlyj

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S.5.pdf
Let $p$ be a real number with $p > 1$. We define $q$ (‘the conjugate exponent’ to $p$) by the equation $1/p + 1/q = 1$. It is easy to check that then $1/(p − 1) = q − 1$, so for positive real numbers $s, t$ we have $s = t^{p−1}$ iff $t = s^{q−1}$.
Proposition S.5.7 (Young's inequality) With $p, q$ as above and any positive real numbers $a, b$
\[
a b \leqslant a^p / p+b^q / q .
\]
Proof We shall show that
\[
a b \leqslant \int_0^a t^{p-1} \mathrm{~d} t+\int_0^b s^{q-1} \mathrm{~d} s=a^p / p+b^q / q .
\]
The right-hand equality comes from elementary integration, while the left-hand inequality is seen in terms of area in the diagram below. The idea is that the first integral is the area shaded vertically while the second is the area shaded horizontally. The graph of $t \mapsto t^{p-1}$ as $t$ goes from 0 to $a$ along the horizontal axis, matches the graph of $s \mapsto s^{q-1}$ as $s$ goes from 0 to $b$ along the vertical axis, because $s=t^{p-1}$ iff $t=s^{q-1}$. The left-hand diagram is for the case $a \leqslant b$ and the right-hand one is for $b \leqslant a$.

As already noted, in each diagram the horizontally shaded area is $\int_0^b t^{q-1} \mathrm{~d} t$ while the vertically shaded area is $\int_0^a s^{p-1} \mathrm{~d} s$. In each case the sum of the shaded areas is at least as big as the area $a b$ of the rectangle with corners $(0,0),(a, 0),(0, b)$ and $(a, b)$. This gives the left-hand inequality at the beginning, and hence completes the proof.

The proof using area is the traditional one. Here is an alternative. Let $f(x)=a x-a^p / p-x^q / q$. Then $f^{\prime}(x)=a-x^{q-1}$ which is zero iff $x=a^{1 /(q-1)}=a^{p-1}$. This is a local maximum since $f^{\prime \prime}(x)=(1-q) x^{q-2}$, and $f^{\prime \prime}\left(a^{p-1}\right)<0$ since $1-q<0$. Now $f\left(a^{p-1}\right)=a^p-a^p / p-a^p / q=0$.
Also, $f(0)<0$, and $f(x) \rightarrow-\infty$ as $x \rightarrow \infty$, since $q>1$. Hence $f(x) \leqslant 0$ for all $x \in[0, \infty)$, and in particular $f(b) \leqslant 0$, which gives Young's inequality.