Integral inequality possibly related to probability theory

Let $$f$$ be Riemann integrable such that $$\int_a^b f(t) \ dt = 1$$ and $$f \geq 0$$ on $$[a,b]$$. If $$\sigma \in C^2$$ is convex, show

$$\sigma\left(\int_a^b tf(t) \ dt\right)\leq\int_a^b f(t)\sigma(t) \ dt$$

and, in addition, discuss when equality holds given the tighter condition $$f>0$$.

I'm assuming this is related to probability theory, since $$f$$ meets the conditions to be a probability density function.

I was able to prove this using Jensen's inequality; however, my professor insists that isn't used. I tried a few other approaches, like fixing one variable and differentiating with respect to the other, but this doesn't seem to lead anywhere. Please help!

• Try to incorporate the proof of Jensen's inequality itself. Mar 8 '17 at 6:18
• Many methods of proving Jensen's is available here (en.m.wikipedia.org/wiki/Jensen's_inequality). Basically one needs to use the convexity of $\sigma$. Mar 8 '17 at 6:22

By integration by parts, $$\int_a^b t f(t) \, \mathrm{d}t = b - \int_a^b F(t) \, \mathrm{d}t$$ where $F(t) = \int_a^t f(s) \, \mathrm{d}x.$
Consider the function $$g(x) = \sigma \Big( x - \int_a^x F(t) \, \mathrm{d}t \Big).$$ It is differentiable with $$g'(x) = (1 - F(x)) \sigma'\Big(x - \int_a^x F(t) \, \mathrm{d}t\Big).$$ Since $\sigma'$ is monotone increasing and $F(t) \ge 0$, we can bound $$g'(x) \le (1 - F(x)) \sigma'(x)$$ and integrating that inequality over $[a,b]$ gives $$\sigma \Big( \int_a^b t f(t) \, \mathrm{d}t \Big) \le \sigma(b) - \int_a^b F(x) \sigma'(x) \, \mathrm{d}x = \int_a^b f(x) \sigma(x) \, \mathrm{d}x,$$ using integration by parts again.