I would like to know if the following inequality is satisfied by all probability distributions (or at least some class of probability distributions) for all integer $n \geq 2$.

$\int_0^{\infty} F(z)^{n-1}(1-\frac{F(z)}{n})\left[zF(z)^{n-2} - \int_0^z F(t)^{n-2}dt\right]f(z)dz$ $\leq \int_0^{\infty} F(z)^{n-1}\left[zF(z)^{n-1} - \int_0^z F(t)^{n-1}dt\right]f(z)dz $

Some comments follow:

1) F(z) is the cumulative distribution function of any probability distribution over positive real numbers. The outer integral runs over the entire support of the distribution, thus, in general, from zero to infinity. f(z) is the probability density function.

2) I will be happy even if this is proved for bounded support distributions, in which case, the outer integral runs from 0 to some upper limit H.

3) Note that both the LHS and the RHS are always non-negative. This is because of the special form of what is inside the square brackets. For both the LHS and the RHS, the second term in the square bracket (i.e. the negative integral from 0 to z), when replaced by its value at the upper limit throughout the region of integration from 0 to z, gives precisely the first term in the square bracket. Thus the term in the square bracket is always non-negative, for both the LHS and the RHS.

4) Also note the obvious similarity in the structure between the LHS and the RHS. The only differences are the difference in exponent for what is inside the square brackets, and the LHS having an extra factor of $1 - \frac{F(z)}{n}$.

5) Note that for $n=2$, this inequality is definitely true since the LHS evaluates to zero owing to the term in the square bracket in the LHS becoming zero, and the RHS is always non-negative, as mentioned in point 3 above.

6) It is easy to work out that for all $n \geq 2$, this inequality is true for uniform [0,1] distributions, i.e., the distribution with support [0,1] and $F(z) = z$.

7) I tried evaluating this integral for small values of $n$ (till 50) using Maple for exponential distribution, and the positive half of the normal distribution and found it to be true. I am guessing that this inequality is true for at least some large class of probability distributions if not all distributions.

8) For instance, would a monotone hazard rate condition help? Monotone hazard rate means that $\frac{f(z)}{1-F(z)}$ is non-decreasing.