<s>[INST] Proving an inequality with n positive real numbers Is there any alternative  way of proving
$$\sum_{k=1}^n\frac{a_k-1}{a_k+1}\geq\frac{\prod_{k=1}^n a_k -1}{\prod_{k=1}^na_k+1}$$ for any positive real numbers $a_1,a_2,\cdots,a_n,$  greater than or equal to 1, other than the one   given in one of the the answers; An inequality associated with sum of n real numbers?
 [/INST] We write the inequality as
$$\sum_{i=1}^n\frac{1}{a_i+1}\leq \frac{n-1}{2}+\frac{1}{a_1a_2\cdots a_n+1}. \quad (1)$$
For $n=1$ the inequality become
$$\frac{1}{a_1+1}+\frac{1}{a_2+1} \leqslant \frac{1}{2} + \frac{1}{a_1a_2+1}, \quad (2)$$
or
$$\frac{(a_1-1)(a_2-1)(a_1a_2-1)}{2(a_1+1)(a_2+1)(a_1a_2+1)} \geqslant 0.$$
Which is true.
Now, let $x=a_1a_2 \cdots a_k,$ suppose $(1)$ is true for $n=k,$ we get
$$\sum_{i=1}^k\frac{1}{a_i+1} \leqslant \frac{k-1}{2}+\frac{1}{x+1}. \quad (3)$$
We will show that this is also true for $n=k+1.$
Indeed, setting $y = a_{k+1},$ use $(3)$ and $(2)$ we have
$$\sum_{i=1}^{k+1}\frac{1}{a_i+1} = \sum_{i=1}^k\frac{1}{a_i+1} + \frac{1}{y+1} \leqslant \frac{k-1}{2}+\frac{1}{x+1} + \frac{1}{y+1}$$
$$\leqslant \frac{k-1}{2}+\frac{1}{2}+\frac{1}{xy+1} = \frac{k}{2}+\frac{1}{x_1x_2\cdots x_{k+1}+1}.$$
Done.
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