I want to prove the negativeness of the following quantity
$$f(x)=v^2e^{-\lambda v} [1-(1+\lambda x)e^{-\lambda x}]-x^2e^{-\lambda x} [1-(1+\lambda v)e^{-\lambda v}],$$where $\lambda$ is a postive constant, $v$ is a positive constant and $x\in[0,v]$.
I plotted this quantity by using some values of $\lambda$ and $v$ in mathematica and it shows that it is always negative, but I want to prove it rigorously.
So I first observe that it is a continuous function, so it has extremes. I calculate $x_0$ such that $f'(x_0)=0$. The result is $$x_0=\frac{{\lambda}^2 v^2+2 {\lambda}v-2 e^{{\lambda} v}+2}{{\lambda} \left({\lambda} v-e^{{\lambda} v}+1\right)}.$$
Although I cannot prove that $f(x_0)$ is always negative, but the graph shows it is.
Anyone could help with proving the negativeness of $f(x$)?
Clearly, $f(0)=f(v)=0.$
Let $A(w) = (w-1)e^{\displaystyle w}+1$ and $G(w) = (w-2)e^{\displaystyle w}+w+2$, for $w \in [0, \infty)$. Then, $A'(w)=we^{\displaystyle w}$. Thus, $A'(w) > 0$, for $w \in (0, \infty)$. Hence, $A(w) > A(0) = 0$, for $w \in (0,\infty)$. $G'(w) = A(w)$. So, $G(w) > G(0) = 0$ for $w \in (0, \infty)$. Let $H(w) = \displaystyle\frac{e^{\displaystyle \lambda w}-1-\lambda w}{w^2}$, for $w \in (0, \infty)$. Then, $H'(w)=\displaystyle\frac{G(\lambda w)}{w^3}>0$. Therefore, $H$ is strictly increasing.
Suppose $x \in (0, v)$. Then, $H(x) < H(v)$. That is, $0 > (xv)^{2}e^{\displaystyle -(x + v)\lambda}(H(x)-H(v)) = f(x).$