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

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.

• For $n=3$ and $\lambda=\frac{2n-1}{(n-1)^2}$ we obtain known Vasc's inequality. In the general, by his HCF Theorem it's enough to prove your inequality for equality case of $n-1$ variables, which gives something very ugly. Commented Nov 7, 2022 at 13:24

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 $$f(1,1,\cdots, 1) = \dfrac{n}{\sqrt{1+\lambda}}$$.

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 $$\mathbb{R}^n_{>0}$$.
• I don't want to waste 1 point for downvoting you, but this is not the solution.
– NN2
Commented Nov 7, 2022 at 9:33
• @NN2 Even without wasting a point, it would be useful to ellaborate. Commented Nov 7, 2022 at 9:39
• It is impossible to solve the system of equations $\frac{\partial L}{\partial x_i} = 0$!
– NN2
Commented Nov 7, 2022 at 9:43
• @NN2 It is not... Every equation will be of the form $$2 \sqrt{1+\lambda x_i} = \frac{\mu}{x_i}$$ It is just a matter of using the restriction to get $\prod_{j\ne i} x_j = \frac{1}{x_i}$ Commented Nov 7, 2022 at 9:45
• But how you solve all $n$ cubic equations with the constraint $\prod x_i = 1$? It's impossible.
– NN2
Commented Nov 7, 2022 at 9:47