Consider:
\begin{equation} \Big(e^x-1\Big)\mathbb{1}_{(x\geq0)} \leq \lambda_1+\lambda_2e^x + \lambda_3x^2 \end{equation} where $\lambda_1$, $\lambda_2$, $\lambda_3$ are three unknown constants. By plotting these two functions, it seems that it is possible to find particular values for $\lambda_i$, $i=1,2,3$, such that \begin{equation} \Big(e^x-1\Big)\mathbb{1}_{(x\geq0)} = \lambda_1+\lambda_2e^x + \lambda_3x^2 \end{equation} at two particular points, i.e. $x_l<0$ and $x_r>0$ are points of tangency.
Could anyone suggest the methods needed to prove this analytically?
The inequality for large $x$ implies that $\lambda_2\geq1.$ The function $f(x)=\lambda_1+\lambda_3x^2$ determines a parabola centered around the $Y$ axis (where in order to cover the case $\lambda_3=0$ we temporarily agree to call a horizontal line a parabola, as well).
Thus for any given $\lambda_2\geq1$ we look for a parabola centered around the $Y$ axis that is tangent to the graph of the function
$$g(x)=\lambda_2e^x-(e^x-1)1_{(x\geq0)}$$
in at least one point to the left of the $Y$ axis and in another point to the right of it.
Such a parabola cannot exist because, being centered around the $Y$ axis, its slope on both sides of the $Y$ axis has opposite sign (or both slopes are zero); whereas the graph of $g$ always has an upward slope (except for $\lambda_2=1$ where the graph is horizontal for positive $x$).