How to find largest $\lambda$ where $1- \lambda^{n}(1-\frac{x}{n})^{n} \geq \lambda x $ holds for all $n$? I am trying to find the largest $\lambda$ where  $$1- \lambda^{n}(1-\frac{x}{n})^{n} \geq \lambda x $$ holds for all integers $n\geq 1$ with $0\leq x\leq 1$. I did a grid search and it is somewhere around $\lambda^* \in [0.82, 0.83]$, but I am not sure if it is possible to calculate it in closed form.
I faced the inequality when proving the approximation ratio of a randomised approximation algorithm. More specifically, it's a sub lemma of Exercise 5.8 in designofapproxalgs.com/book.pdf.

So you have the good guess.

By choosing $n=2$ and $x=1$, we see that $\lambda$ has to be smaller than or equal to $2(\sqrt{2}-1)$. 
Now we will prove that $\lambda=2(\sqrt{2}-1)$ is also a value for which your initial inequality is true.
Firstly, by our choice of $\lambda$, we see that the following inequality is true for $n=1,2$
$$1 \ge \lambda + \lambda^n(1-\frac{1}{n})^n \quad (*)$$
For $n \ge 2$, you see that indeed, $n \mapsto  \lambda^n(1-\frac{1}{n})^n$ is a decreasing with $\lambda =2(\sqrt{2}-1)$. So because $(*)$ is true for $n=2$, we imply that it is true for all $n$.
Now in the last step, to prove that $1 \ge \lambda x + \lambda^n(1-\frac{x}{n})^n $ for all $n \ge 1$ and $x \in [0,1]$, you can use the observation that the RHS is a convex function on $x$, so its maximum is achieved at either $x=0$ or $x=1$. But,
$$RHS_{x=0}= \lambda^n \le 1 \quad \text{(true)}$$
and $$RHS_{x=1} = \lambda+(1-\frac{1}{n})^n\lambda^n \le 1 \quad \text{(true)}$$
So in conclusion, $\lambda= 2(\sqrt{2}-1)$ is the maximum value of $\lambda$ such that :
$$1 \ge \lambda x + \lambda^n(1-\frac{x}{n})^n $$ for all $n \ge 1$ and $x \in [0,1]$
