Prove that $\int_{X>\alpha}XdP\le\alpha(-\log F(\alpha))+\int_\alpha^\infty(-\log F(t))\lambda(dt)$

Question

Prove that $$\int_{X>\alpha}XdP\le\alpha(-\log F(\alpha))+\int_\alpha^\infty(-\log F(t))\lambda(dt)\le\frac{1}{F(\alpha)}\int_{X>\alpha}XdP$$

While attempting to do this, using basic substitution and Fubini's, I ended up with $$\int_{X\alpha}XdP=\alpha(1-F(\alpha))+\int_\alpha^\infty(1-F(t))\lambda(dt)$$ Now, however using the fact that $F(\alpha)\in[0,1]$, is it possible to show that $F(\alpha)-\log F(\alpha)\ge1$ which would complete the first inequality. $x-\log x$ is a decreasing function and at $x=1$, it is exactly equal to $1$. Is this correct? What about the second inequality? Any help?

Answer 1

Hint for the both inequalities:

The $\log$ is a concave function so:

$$\log x \le \log 1 + \log'(1) (x-1).$$

For the second inequality,

\begin{align} \alpha(-\log F(\alpha))+\int_\alpha^\infty(-\log F(t))\lambda(dt) &= \alpha\log\frac{1}{F(\alpha)} + \int_\alpha^\infty \log \frac1{F(t)}\lambda(dt)\\ &\le \alpha \left(\frac1{F(\alpha)} - 1\right) + \int_{\alpha}^{\infty}\left(\frac1{F(t)} - 1\right)\lambda(dt)\\ &\le \frac1{F(\alpha)} \left(\alpha \left(1 -F(\alpha)\right) + \int_\alpha^\infty (1-F(t))\lambda(dt)\right). \end{align}

Answer 2

Hint for the second inequality,

\begin{align} \alpha(-\log F(\alpha))+\int_\alpha^\infty(-\log F(t))\lambda(dt) &= -\alpha \log F(\alpha) - \left[t\log t - t\right]_{F(\alpha)}^{1}\\ &= -(\alpha -F(\alpha))\log F(\alpha) + 1 - F(\alpha) \end{align}