Let $f: [0,\infty)\to \mathbb{R}$ a non-decreasing function. Then show this inequality holds for all $x,y,z$ such that $0\le x<y<z$. \begin{align*} & (z-x)\int_{y}^{z}f(u)\,\mathrm{du}\ge (z-y)\int_{x}^{z}f(u)\,\mathrm{du} \end{align*}
Can anyone help me?
Today in the morning I was trying out this problem and I was able to do up until this point before getting stuck. Please help me proceed further.
Here is my attempt at a solution.
Given \begin{align*} & (z-x)\int_{y}^{z}f(u)\,\mathrm{du}\ge (z-y)\int_{x}^{z}f(u)\,\mathrm{du} \end{align*}\begin{align*} & \Longrightarrow(z-x)\int_{y}^{z}f(u)\,\mathrm{du}\ge (z-y)\int_{x}^{y}f(u)+(z-y)\int_{y}^{z}f(u)\,\mathrm{du} \end{align*}\begin{align*} & \Longrightarrow(y-x)\int_{y}^{z}f(u)\,\mathrm{du}\ge (z-y)\int_{x}^{y}f(u)\,\mathrm{du} \end{align*}\begin{align*} & \Longrightarrow(y-x)(f(z)-f(y))\,\mathrm\ge (z-y)(f(y)-f(x))\,\mathrm\ \end{align*}\begin{align*} & \Longrightarrow(y-x)f(z)+(z-y)f(x)\,\mathrm\ge (z-x)f(y)\mathrm\ \end{align*} Note: If my approach is wrong or inefficient, please let me know: I'd be happy to learn new methods.
You shouldn't start with what you're trying to prove. Note that the result you're trying to prove is intuitively obvious if you divide both sides by $(z-y)(z-x)$ : The average value over $[y,z]$ is larger than the average value over $[x,z]$ because the function is bigger over $[x,y]$ than $[y,z]$. Basically this is like saying the average of 2, 4, and 5 is bigger than the average of 1, 2, 4 and 5.
Try proving that the average of 3 numbers $b\leq c\leq d$ is greater than the average of the 4 numbers $a\leq b \leq c \leq d$. Then you can try to generalize this result to integrals by breaking up the interval you're integrating over into two pieces.