If H(b) $\leq$ b then $\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b) - [1-b]db $? $H$ is a CDF over $[0,1]$
I want to prove that if $H(b) \leq b$ then :
$$\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b) - [1-b]db $$ for all $k \in [0,1]$
I think this is true, but a rigorous proof does seem difficult to me. Can anyone help me to prove this? or does any theorem helps?

I assume $h(b)=H'(b)$ is a density, $v$ is a constant and $H(b) \leq b \quad\forall b$
\begin{align}
&H(b) \leq b \implies H(b)-1 \leq b-1 \implies 1-H(b)\geq1-b \implies -(1-H(b))\leq-(1-b) 
\end{align}
Substituting in the equation you get the following inequality.
$\int_0^k (v-b)h(b) - [1-H(b)]db \leq \int_0^k (v-b)h(b) - [1-b]db$
Now we need to show that substituting $h(b)$ for $1$ gives us the desired inequality.
We know that $\int_0^k(v-b)db = vk -\frac{k^2}{2}$ then:
\begin{align}
\int_0^k (v-b)h(b)db &= v\int_0^kh(b)db-\int_0^kbh(b)db\\
&= v[H(k)-H(0)]-\int_0^kbh(b)db\\
&= v(H(k)-H(0))-bH(b)|_0^k +\int_0^kH(b)db\\
&\leq(v-k)H(k)-H(0)v +\int_0^kbdb\\
&H(b)\leq b\; \forall b\; \wedge \text{H(.) is a cdf}\implies H(0)=0 \\
&\leq (v-k)k+\frac{k^2}{2}\\
&=vk-\frac{k^2}{2}
\end{align}
then the following holds whenever $v \geq k$
$$\int_0^k (v-b)h(b)db \leq vk-\frac{k^2}{2} = \int_0^k(v-b)db$$
