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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Nov 7, 2022 at 13:24

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

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