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?

It is obvious that 
$$\int_0^k ((v-b)h(b) - (1-H(b)))db \leq \int_0^k ((v-b) - (1-b))db$$ if and only if
$$\int_0^k ((v-b)h(b) + H(b))db \leq \int_0^k ((v-b) +b)db$$ or equivalently
$$\int_0^k ((v-b)h(b) + H(b))db \leq \int_0^k vdb.$$
We will work with the LHS. We have
$$\int_0^k ((v-b)h(b) + H(b))db =vH(k)+\int_0^k(H(b)-bh(b))db.$$
Now, using integration by parts, it is 
$$\int_0^kbh(b)db=[bH(b)]_0^k-\int_0^kH(b)db=kH(k)-\int_0^kH(b)db.$$ Thus
$$\int_0^k ((v-b)h(b) + H(b))db =vH(k)-kH(k).$$ On the other hand, the RHS is 
$$\int_0^k vdb=vk.$$ So, the given inequality is equivalent to 
$$vH(k)-kH(k)\le kv.$$ In other words
$$-kH(k)\le v(k-H(k)).$$ Now we see that
$$-kH(k)\le 0 \le v(k-k)\le v(k-H(k)),$$ where we have used $H(k)\le k \implies-k\le -H(k).$
