Let $f:[0,\infty)\to\mathbb{R}$ be a non-decreasing function. Then $(z-x)\int_y^zf(u)du\ge (z-y)\int_x^zf(u)du$ , where $0\le xLet $f:[0,\infty)\to\mathbb{R}$ be a non-decreasing function. Then show that the following inequality holds for all $x,y,z$ such that $0\le x<y<z$ $$(z-x)\int_y^zf(u)du\ge (z-y)\int_x^zf(u)du.$$ 
My approach: Since, $0\le x<y<z$ and $f$ is non-decreasing, we have $f(0)\le f(x)\le f(y)\le f(z)$. 
Now, $(z-x)\int_y^zf(u)du=(z-y+y-x)\int_y^zf(u)du=(z-y)\int_y^zf(u)du+(y-x)\int_y^zf(u)du.$ 
Again, $(z-y)\int_x^zf(u)du=(z-y)[\int_x^yf(u)du+\int_y^zf(u)du].$
Now, $$(z-x)\int_y^zf(u)du\ge (z-y)\int_x^zf(u)du$$ $$\Leftrightarrow (z-y)\int_y^zf(u)du+(y-x)\int_y^zf(u)du\ge (z-y)\int_x^yf(u)du+(z-y)\int_y^zf(u)du$$ $$\Leftrightarrow (y-x)\int_y^zf(u)du\ge (z-y)\int_x^yf(u)du$$ 
Therefore, if we can prove that $$(y-x)\int_y^zf(u)du\ge (z-y)\int_x^yf(u)du,$$ we will be done. 
 A: Okay, here goes the solution. 
It is given that $f$ is non-decreasing on $[0,\infty)$ and $0\le x<y<z.$ Since, $f$ is non-decreasing on $[0,\infty)$, implies that $f$ is monotone on $[0,\infty)$. Therefore, $f$ is integrable on $[0,\infty)$.  
Now, $(z-x)\int_y^zf(u)du=(z-y+y-x)\int_y^zf(u)du=(z-y)\int_y^zf(u)du+(y-x)\int_y^zf(u)du.$ 
Again, $(z-y)\int_x^zf(u)du=(z-y)[\int_x^yf(u)du+\int_y^zf(u)du].$
Now, $$(z-x)\int_y^zf(u)du\ge (z-y)\int_x^zf(u)du$$ $$\Leftrightarrow (z-y)\int_y^zf(u)du+(y-x)\int_y^zf(u)du\ge (z-y)\int_x^yf(u)du+(z-y)\int_y^zf(u)du$$ $$\Leftrightarrow (y-x)\int_y^zf(u)du\ge (z-y)\int_x^yf(u)du.$$ 
Therefore, if we show that $$(y-x)\int_y^zf(u)du\ge (z-y)\int_x^yf(u)du$$ holds true, we will be done. We proceed towards proving the same. 
We know that if a function $f$ is integrable on $[a,b]$, and $m=\inf\{f(x)|a\le x\le b\}$ and $M=\sup\{f(x)|a\le x\le b\}$, then $$m(b-a)\le \int_a^b f(x)dx\le M(b-a). (*)$$ 
Now consider the interval $S_1=[x,y]$. For $S_1$, $m=f(x)$ and $M=f(y)$ (since, $f$ is non-decreasing on $S_1$). Then using $(*)$, we can say that $$f(x)(y-x)\le\int_x^yf(u)du\le f(y)(y-x)$$ $$\implies f(x)\le\frac{\int_x^yf(u)du}{y-x}\le f(y). (1)$$
Now consider the interval $S_2=[y,z]$. For $S_2$, $m=f(y)$ and $M=f(z)$ (since, $f$ is non-decreasing on $S_2$). Then using $(*)$, we can say that $$f(y)(z-y)\le\int_y^zf(u)du\le f(z)(z-y)$$ $$\implies f(y)\le\frac{\int_y^zf(u)du}{z-y}\le f(z). (2)$$
Combining $(1)$ and $(2)$, we get that $$f(x)\le\frac{\int_x^yf(u)du}{y-x}\le f(y)\le\frac{\int_y^zf(u)du}{z-y}\le f(z)$$ $$\implies \frac{\int_x^yf(u)du}{y-x}\le \frac{\int_y^zf(u)du}{z-y}$$ $$\implies (z-y)\int_x^y f(u)du\le (y-x)\int_y^z f(u)du$$ $$\implies (y-x)\int_y^z f(u)du\ge (z-y)\int_x^y f(u)du.$$
Thus, we are done. 
A: Let $$\int f(u)du=F(u)+C$$
Now, using LMVT on F(u) on the interval [y,z], we get
$$\frac{F(z)-F(y)}{z-y}=F'(a), a \in (y,z)$$
$$\Rightarrow \frac{\int_y^z f(u)du}{z-y}=f(a), a\in(y,z)$$
Similarly,$$\frac{\int_x^y f(u)du}{y-x}=f(b), b\in(x,y)$$
Now,
f(u) is non decreasing.
So, $$f(a)\ge f(y) \ge f(b)$$
$$\Rightarrow \frac{\int_y^z f(u)du}{z-y} \ge \frac{\int_x^y f(u)du}{y-x}$$
$$\Rightarrow(y-x)\int_y^z f(u)du \ge (z-y)\int_x^y f(u)du$$
$$\Rightarrow(y-x)\int_y^z f(u)du + (z-y)\int_y^z f(u)du \ge (z-y)\int_x^y f(u)du + (z-y)\int_y^z f(u)du$$
$$\Rightarrow(z-x)\int_y^z f(u)du \ge (z-y)\int_x^z f(u)du$$
Hence proved.
Hope this will help you.
A: Hint:
You need to prove that $$(z-x)\int_y^zf(u)du\ge (z-y)\int_x^z f(u)du$$
I.e you need to prove that $$\displaystyle \frac{\displaystyle \int_y^zf(u)du}{\displaystyle \int_y^z du}\ge \frac{\displaystyle \int_x^zf(u)du}{\displaystyle\int_x^zdu}$$
Now can you relate this inequality with the "average" of $f$ on those intervals using that '$f$' is non-decreasing?
