The chapter of this problem is Solving Elementary Inequality Using Integrals. After I typed the problem: I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.
For part $a$,I can prove AM-GM inequality by intgeral,But for Weighted AM–GM inequality I can't
My approach maybe
$$\dfrac{A}{G}-1=\sum_{i=1}^{k} r_i \int_{a_i}^{G} (\frac{1}{t} – \frac{1}{G}) dt + \sum_{i=k+1}^{n} r_i \int_{G}^{a_i} (\frac{1}{G} – \frac{1}{t}) dt $$
For part $b$: I can solve it(without use integral),because it must to prove
$$\dfrac{A}{A'}\ge\dfrac{G}{G'}$$ or
$$\dfrac{\sum_{i=1}^{n}r_{i}x_{i}}{\sum_{i=1}^{n}r_{i}(1-x_{i})}\ge \prod_{i=1}^{n}\left(\dfrac{x_{i}}{1-x_{i}}\right)^{r_{i}}$$
so we consider
$$f(x)=\ln{\dfrac{x}{1-x}}=\ln{x}-\ln{(1-x)}(0<x<0.5)$$
and
$$f'(x)=\dfrac{1}{x}+\dfrac{1}{1-x},f''(x)=-\dfrac{1}{x^2}+\dfrac{1}{(1-x)^2}<0(0<x<\frac{1}{2}$$
so By Jenson inequality
$$\sum_{i=1}^{n}r_{i}f(x_{i})\le f(\sum_{i=1}^{n}r_{i}x_{i})$$
so I have prove it part b,
But I can't prove part a
Best Answer
$\newcommand{\d}{\,\mathrm{d}}$The first step is in evaluating $I$. Let's first note that each $x_i$ and $A$ itself are strictly positive so we only care about evaluating $I$ in the case $x,a\gt0$. A partial fraction decomposition is in order - I leave the derivation at the bottom of the post for readability:
$$\frac{t}{(1+t)(x+at)^2}=\frac{x}{x-a}\cdot\frac{1}{(x+at)^2}+\frac{a}{(x-a)^2}\cdot\frac{1}{x+at}-\frac{1}{(x-a)^2}\cdot\frac{1}{1+t}$$
So:
$$\begin{align}I(x,a)&=\frac{x}{x-a}\int_0^\infty\frac{1}{(x+at)^2}\d t+\frac{1}{(x-a)^2}\int_0^\infty\left(\frac{a}{x+at}-\frac{1}{1+t}\right)\d t\\&\overset{u=at}{=}\frac{x}{x-a}\cdot\frac{1}{a}\int_0^\infty\frac{1}{(x+t)^2}\d t+\frac{1}{(x-a)^2}[\ln(x+at)-\ln(1+t)]_0^\infty\\&\overset{u=t+x}{=}\frac{x}{a(x-a)}\int_x^\infty\frac{1}{t^2}\d t+\frac{1}{(x-a)^2}\left[\lim_{t\to\infty}\ln\left(\frac{x+at}{1+t}\right)-\ln x\right]\\&=\frac{1}{a(x-a)}+\frac{1}{(x-a)^2}[\ln a-\ln x]\end{align}$$
Let's turn our attention to the summand: $$\begin{align}r_i(x_i-A)^2I(x_i,A)&=r_i(x_i-A)^2\left(\frac{1}{A(x_i-A)}+\frac{1}{(x_i-A)^2}\ln A-\frac{1}{(x_i-A)^2}\ln x_i\right)\\&=r_i\left(\frac{x_i}{A}-\frac{A}{A}+\ln A-\ln x_i\right)\\&=-r_i+\frac{1}{A}r_ix_i+\ln A\cdot r_i-r_i\ln x_i\end{align}$$
And let's sum, recalling that $\sum_{i=1}^nr_i=1$:
$$\begin{align}\sum_{i=1}^nr_i(x_i-A)^2I(x_i,A)&=-\sum_{i=1}^nr_i+\frac{1}{A}\sum_{i=1}^nr_ix_i+\ln A\cdot\sum_{i=1}^nr_i-\sum_{i=1}^nr_i\ln x_i\\&=-1+\frac{1}{A}(A)+\ln A-\sum_{i=1}^n\ln(x_i^{r_i})\\&=\ln A-\ln\left(\prod_{i=1}^nx_i^{r_i}\right)\\&=\ln\frac{A}{G}\end{align}$$
How do we conclude? Well, as $r_i,x_i,A\gt0$ always, the integral $I(x_i,A)$ is an integral of purely nonnegative terms and is thus nonnegative; $(x_i-A)^2\ge0$ since the squares of real numbers are nonnegative, and as said $r_i\gt0$ always, so each summand $r_i(x_i-A)^2I(x_i,A)\ge0$ for all $i$, so that: $$\begin{align}\ln\frac{A}{G}&=\sum_{i=1}^nr_i(x_i-A)^2I(x_i,A)\ge0\\\ln\frac{A}{G}&\ge0\\\frac{A}{G}&\ge1\\A&\ge G\quad\blacksquare\end{align}$$
And we are all done!
N.B. On the partial fraction decomposition: