concave function

Hallo!
I've to show that the function $Tv$ is concave in $x$.

$Tv: \mathbb{R}_+ \to \mathbb{R}$ defined by
$Tv(x):=\sup\limits_{a \in [0,1]}\mathbb{E}[v(e^{rh}(1+a(R-1)))]$
where
$v:\mathbb{R}_+ \to \mathbb{R}$ is a continous function with the follwing properties
$v(x)\leq c(1+x)$ for a constant $c \geq 0$,
strictly concave and strictly increasing.

Furthermore
$\mathbb{E}$ expectation,
$R$ lognormal distributed random variable,
$r,h>0$ are constants

Let $\lambda \in (0,1)$ and $x_1, x_2 \in \mathbb{R}_+.$ $v$ is concave, so it holds
$Tv(\lambda x_1+ (1-\lambda)x_2)\geq \sup\limits_{a \in [0,1]}{(\lambda \mathbb{E}v(e^{rh}x_1(1+a( R-1))+(1-\lambda)\mathbb{E}v(e^{rh}x_2(1+a( R-1)))$

But why is
$Tv(\lambda x_1+ (1-\lambda)x_2)\geq \lambda Tv(x_1)+(1-\lambda) Tv(x_2)$?

Can anybody help me?