$\frac{1}{\sqrt{1+\lambda x_1}}+\frac{1}{\sqrt{1+\lambda x_2}}+...+\frac{1}{\sqrt{1+\lambda x_n}}\leq \frac{n}{\sqrt{1+\lambda}}$

Question

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.

Answer 1

The maximisers of $\displaystyle f(x) = \sum_{i=1}^n\dfrac{1}{\sqrt{1+\lambda x_i}}$ will be stationary points of the Lagrangian $$ L(x_1, \cdots, x_n; \mu) = \sum_{i=1}^n\dfrac{1}{\sqrt{1+\lambda x_i}} - \mu x_1 \cdots x_n $$

Solving the corresponding system, you'll see that the maximizer will satisfy $x_1=x_2 = \cdots = x_n$, which means that all $x_i$'s must be one in order to maximise $f$. The proof is "complete" by noting that $f(1,1,\cdots, 1) = \dfrac{n}{\sqrt{1+\lambda}}$.

Still needs to be addressed:

1. Using second order conditions to show that this single stationary point of the Lagrangian is indeed a maximiser.
2. Showing that the objective function does not extend to the boundary with a value greater than $f(1,\cdots,1)$, making the stationary point a global maximiser in $\mathbb{R}^n_{>0}$.