Let $x_i\geq 1, 1\leq i\le n.$ Then show that $$\sum_{i=1}^n\frac{1}{1+x_i}\leq \frac{n-1}{2}+\frac{1}{1+x_1x_2\cdots x_n}.$$ One method of proving this is induction on $n.$ This can be achieved by the fact; for $n=2,$ $ \frac{1}{1+x_1}+\frac{1}{1+x_2}\leq \frac{1}{2}+\frac{1}{1+x_1x_2}.$ This idea can be extended to complete the proof of induction.
I am curious to know if you have any alternative attractive proofs! Solutions using multivariable calculus techniques are also welcome.
Possibility of sharpening; I was wondering if the bound can be sharpened further, especially with terms containing symmetric functions of $x_1,.....x_n?$
Your new inequality we can prove by Karamata for the concave on $[0,+\infty)$ function $f(x)=\frac{e^x-1}{e^x+1}$.
Indeed, for $x\geq0$ we obtain: $$f''(x)=\frac{2e^x(1-e^x)}{(e^x+1)^3}\leq0.$$ Now, let $x_i=e^{a_i}.$
Thus, $a_i\geq0$ and we need to prove that $$\sum_{i=1}^n\left(\frac{1}{1+x_i}-\frac{1}{2}\right)\leq\frac{1}{1+x_1...x_n}-\frac{1}{2}$$ or $$\sum_{i=1}^n\frac{x_i-1}{x_i+1}\geq\frac{x_1...x_n-1}{x_1...x_n+1}.$$ Now, let $x_1\geq x_2\geq...\geq x_n.$
Thus, $$(a_1+a_2+...+a_n,0,...,0)\succ(a_1,a_2,...a_n),$$ which by Karamata gives:
$$\sum_{i=1}^n\frac{x_i-1}{x_i+1}=\sum_{i=1}^n\frac{e^{a_i}-1}{e^{a_i}+1}\geq\frac{e^{a_1+...+a_n}-1}{e^{a_1+...+a_n}+1}=\frac{x_1...x_n-1}{x_1...x_n+1}$$ and we are done!