Suppose we have $$\frac1n\sum_{i=1}^n\frac{g_i}{1+\lambda g_i}=0.$$ Since $\frac{1}{1+ax}=1-\frac{ax}{(1+a\xi)^2}$, $\xi$ is between $0$ and $x$, we obtain $$\frac1n\sum_{i=1}^n\frac{g_i}{1+\lambda g_i}=\frac1n\sum_{i=1}^ng_i\cdot\Big[1-\frac{\lambda g_i}{(1+\tilde\lambda g_i)^2}\Big]=\frac1n\sum_{i=1}^ng_i-\lambda\cdot\frac1n\sum_{i=1}^n\frac{g_i^2}{(1+\tilde\lambda g_i)^2}=0,$$ where $\tilde\lambda$ is between $0$ and $\lambda$. Denote $\hat V_1=\frac1n\sum_{i=1}^n\frac{g_i^2}{(1+\tilde\lambda g_i)^2}$, we have $$\lambda=\hat V_1^{-1}\cdot\frac1n\sum_{i=1}^ng_i$$
Suppose $\frac1n\sum_{i=1}^ng_i^2\overset{p}\to V$ and $\max_{1\le i\le n}|\lambda g_i|\overset{p}\to 0$. I saw in one paper that $$|\hat V_1-V|\le \max_{1\le i\le n}\Big|\frac{1}{1+\tilde\lambda g_i}\Big|^2\cdot|\frac1n\sum_{i=1}^ng_i^2- V|\overset{p}\to0$$
My question is that how does this inequality hold? Does it have any trick?
If we use the approximation $\frac{1}{1+ax}\approx 1-ax$, we chave $$\frac1n\sum_{i=1}^n\frac{g_i}{1+\lambda g_i}=\frac1n\sum_{i=1}^ng_i\cdot(1-\lambda g_i)=\frac1n\sum_{i=1}^ng_i-\lambda\cdot\frac1n\sum_{i=1}^ng_i^2=0,$$ Denote $\hat V_2=\frac1n\sum_{i=1}^ng_i^2$, under the assumption, we can directly obtain that $$\lambda=\hat V_2^{-1}\cdot\frac1n\sum_{i=1}^ng_i ~~\mbox{and}~~ \hat V_2\overset{p}\to V.$$
Is this observation equavalent to the inequality above?