If $f'(x)\ge0$. Prove $${1\over z-y}\int_y^zf(u)\,du\ge{1\over z-x}\int_x^zf(u)\,du$$ $\forall\,\,0\le x<y<z$
Since each $f(u_1)\ge f(u_2)\forall\,\,\,u_1>u_2$ This looks obviously true. Any hints on how to get started for proof?
2026-04-03 04:41:04.1775191264
Proving average area under curve is greater for higher bounds
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Make the substitution $t=\frac {u-y} {z-y}$ in the LHS to get $\int_0^{1} f(tz+(1-t)y)dt$. Similarly, RHS =$\int_0^{1} f(tz+(1-t)x)dt$. Since $f(tz+(1-t)y) \geq f(tz+(1-t)x)$ we are done.