Suppose $G$ is differentiable, increasing, and strictly convex.
As part of a larger problem in microeconomics, I'm trying to prove that if $\lambda>0$, then
$$G'(b_2)+2\lambda G(b_2)>G'(b_1)+\lambda G(b_1)$$
implies $b_2>b_1$.
The converse is obviously true: if $b_2>b_1$, then $G(b_2)>G(b_1)$ because $G$ is increasing, and $G'(b_2)>G'(b_1)$ because $G$ is strictly convex. I'm having trouble with the other direction because of the discrepancy of $\lambda$ versus $2\lambda$.
The result is not true in general. Let $G(x)=e^x$, $\lambda=1$, $b_2=x\in\Bbb R$ and $b_1=y=x+\epsilon$ for $\epsilon>0$ such that $2\,e^\epsilon<3$. Then $x<y$, but \begin{align} G'(x)+2\,\lambda\,G(x)&=3\,e^x\\ &>2\,e^\epsilon e^x\\ &=G'(y)+\lambda\,G(y). \end{align}