Showing $\mathrm{Var}(\min(X,y))$ is increasing in $y$ where $X$ is a random variable For nonnegative random variables, expectation is defined to be the supremum of all expectations of simple random variables $A$ that satisfy $A\leq X$. (For  simple random variables, $E(A)=\sum_jc_jP(C_j)$ where $c_j$'s are nonnegative constants and $C_j$'s are disjoint events in $\Omega$.)
Let $X$ be a nonnegative real-valued random variable, how can we show that  $\mathrm{Var}(\min(X,y))$ is increasing in $y$? ($y$ is a constant.)
I thought that the proof is probably based on definition of expectation and I would need compare $\mathrm{Var}(\min(X,y_1))$ and $\mathrm{Var}(\min(X,y_2))$ for $y_1\leq y_2$ and probably start from some simple RVs. 
Thank you for any help!

Following the hint from @Michael:
$Var(min(X,y))$
= $Var(Z)$
= $\int_0^{\infty} P[Z^2 > t] dt - (\int_0^{\infty} P[Z > t] dt)^2$
= $\int_0^{\infty} P[min(X,y)^2 > t] dt - (\int_0^{\infty} P[min(X,y) > t] dt)^2$
= $\int_0^{y^2} P[X^2 > t] dt - (\int_0^{y} P[X > t] dt)^2$  
Then,
$\frac{d}{dy}[Var(min(X,y))]$
= $ 2y P[X^2 > y^2] - 2 (\int_0^{y} P[X > t] dt) P[X > y]$
= $ 2y P[X > y] - 2 (\int_0^{y} P[X > t] dt) P[X > y]$
= $ 2 P[X > y] (y - \int_0^{y} P[X > t] dt)$
$\geq 0$  
with the last line following from $\int_0^{y} P[X > t] dt \leq \int_0^{y} dt = y$
The inequality must be weak because, if $X$ were a constant then $\forall y: Var(min(X,y)) = 0$ so $Var(min(X,y))$ does not increase in $y$. 
