Let $x_1, x_2,...,x_n$ be positive numbers such that $x_1x_2...x_n=1$.
For $n \geq 3$ and $0<\lambda\leq\frac{2n-1}{(n-1)^2}$, prove that $$\frac{1}{\sqrt{1+\lambda x_1}}+\frac{1}{\sqrt{1+\lambda x_2}}+...+\frac{1}{\sqrt{1+\lambda x_n}}\leq \frac{n}{\sqrt{1+\lambda}}$$
My initial thought: What if I let $v_i=\lambda x_i$, so the given condition will become $$\prod_{i=1}^{n}v_i=\lambda^n$$ and $$\lambda=\left (\prod_{i=1}^{n}v_i\right)^{1/n}\leq \frac{2n-1}{(n-1)^2}$$
$$\therefore \frac{1}{\sqrt{1+\lambda x_1}}+\frac{1}{\sqrt{1+\lambda x_2}}+...+\frac{1}{\sqrt{1+\lambda x_n}}=\sum_{i=1}^n\frac{1}{\sqrt{1+v_i}}\leq \frac{n}{\sqrt{1+\lambda}}$$
But I have no idea how to proceed from here. Any alternative solutions/tricks/guidances are appreciated. Thank you in advance.