# Given positive real numbers $x_1,x_2,\ldots,x_n$ such that $\sum_{k=1}^{n}{\frac{1}{x_k+1}}=n-2$ , how to prove the following inequality?

Given positive real numbers $$x_1,x_2,\ldots,x_n$$ such that $$\sum_{k=1}^{n}{\frac{1}{x_k+1}}=n-2$$ , prove that

$$\sum_{i=1}^{n}{\frac{1}{x_i}}+\frac{n(n-2)}{\sum_{i=1}^{n}{x_i}}\geqslant(n-1)(n-2)$$

This problem was initially asked on the site 'zhihu'. The case $$n=3$$ has been proved by 'SOS' method, and I've given a $$pqr$$ method proof. How to prove (or falsify) this generalised inequality?

I've tried to set $$a_i=1/x_i$$ , then $$\sum_{k=1}^{n}{\frac{a_k}{a_k+1}}=n-2$$ and the inequality is equivalent to

$$nA+(n-2)H\geqslant(n-1)(n-2)$$

where $$A$$ is the Arithmetic Mean among $$a_i$$ , and H is the Harmonic Mean among $$a_i$$ , and maybe we can apply some theorems relevant to these means.

• It was wrong, sorry. Jan 16 at 7:49
• @O-17 I solved it for $n=4$. It's true. Jan 16 at 18:14

Let $$\sum_{i=1}^{n}{\frac{1}{x_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{x_i}}<(n-1)(n-2)$$ for some values of $$x_i$$, $$x_i=ka_i,$$ where $$k>0$$ and $$\sum_{i=1}^{n}{\frac{1}{a_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{a_i}}=(n-1)(n-2).$$ Thus, $$\frac{1}{k}(n-1)(n-2)<(n-1)(n-2),$$ which gives $$k>1$$ and we obtain: $$n-2=\sum_{i=1}^{n}{\frac{1}{x_i+1}}=\sum_{k=1}^{n}{\frac{1}{ka_i+1}}<\sum_{k=1}^{n}{\frac{1}{a_k+1}},$$ which is a contradiction because we'll prove now that $$\sum_{k=1}^{n}{\frac{1}{a_k+1}}\leq n-2$$ for any positives $$a_i$$ such that $$\sum_{i=1}^{n}{\frac{1}{a_i}}+\frac{n(n-2)}{\sum\limits_{i=1}^{n}{a_i}}=(n-1)(n-2).$$ Indeed, let $$f(x)=\frac{1}{1+x}$$ and $$g(x)=f'\left(\frac{1}{\sqrt{x}}\right)=-\frac{x}{(1+\sqrt{x})^2}.$$
Thus, $$g''(x)=\frac{3}{2\sqrt{x}(1+\sqrt{x})^4}>0,$$ which says that $$g$$ is a strictly convex on $$(0,+\infty).$$
Now, let $$\sum\limits_{i=1}^na_i=constant$$.
Thus, $$\sum\limits_{i=1}^n\frac{1}{a_i}=constant$$ and by the Vasc's EV Method Corollary 1.6 it's enough to prove that $$\sum\limits_{i=1}^nf(a_i)\leq n-2$$ for equality case of $$n-1$$ variables.
Id est, it's enough to prove that: $$\frac{n-1}{x+1}+\frac{1}{y+1}\leq n-2,$$ where $$x$$ and $$y$$ are positives such that $$\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}=(n-1)(n-2)$$ or
$$\frac{n-1}{x+\frac{(n-1)(n-2)}{\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}}}+\frac{1}{y+\frac{(n-1)(n-2)}{\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}}}\leq\frac{(n-2)\left(\frac{n-1}{x}+\frac{1}{y}+\frac{n(n-2)}{(n-1)x+y}\right)}{(n-1)(n-2)}$$ or $$(x-y)^2x(2nxy+(x-y)^2)\geq0$$ and we are done!