# $\int_0^1f(x)\cdot x^{n+1}\text{d}x > \int_0^1f(x)\cdot x^n\text{d}x \cdot \int_0^1f(x)\cdot x\text{d}x$

I have convinced myself that $$\int_0^1f(x)\cdot x^{n+1}\text{d}x > \int_0^1f(x)\cdot x^n\text{d}x \cdot \int_0^1f(x)\cdot x\text{d}x$$ is true whenever

• $$f$$ is non-negative,
• $$\int_0^1f(x)\text{d}x=1$$, and
• it is not the case that $$f(x)=\delta(x-c)$$, where $$\delta$$ is the Dirac delta function and $$c$$ is a constant in $$[0,1]$$ (in which case it is obvious that equality holds instead).

However, I could use some help proving it. I would be very happy just to get a hint - no full solution needed.

Btw, if there's someone who think they can help but don't know what the Dirac delta function is, then just assume that $$f$$ is a normal function and ignore the last condition.

• – Martin R Mar 5 at 16:29

Presumably $$n > 0$$ here.
This is essentially a special case of the Harris inequality for the probability measure $$f(x)\; dx$$ on $$[0,1]$$. For any strictly increasing functions $$g$$ and $$h$$ we have
$$\int_0^1 f(x) g(x) h(x)\; dx > \int_0^1 f(x) g(x)\; dx \cdot \int_0^1 f(x) h(x)\; dx$$ which follows from $$\int_0^1 \int_0^1 (g(x)-g(y))(h(x)-h(y)) f(x) f(y)\; dx \; dy > 0$$