Do you think I will get any points for that?: I did substitution $sin a_i=x_i$ and wrote $$\Bigg(\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+x_i}\Bigg)\Bigg(1+\prod_{i=1}^{n}(x_i)^{\frac{1}{n}}\Bigg)= \Bigg(\frac{1}{n}\sum_{i=1}^{n}\frac{1}{\frac{1}{n-1}+\cdots+\frac{1}{n-1}+x_i}\Bigg)\Bigg(1+\prod_{i=1}^{n}(x_i)^{\frac{1}{n}}\Bigg)\leq$$ $$\Bigg(\frac{1}{n}\sum_{i=1}^{n}\frac{1}{n\sqrt[n]{\frac{x_i}{(n-1)^{n-1}}}}\Bigg)\Bigg(1+\prod_{i=1}^{n}(x_i)^{\frac{1}{n}}\Bigg)=$$ $$\frac{\sqrt[n]{(n-1)^{n-1}}}{n^2}\Bigg(\sum_{i=1}^{n}\frac{1}{\sqrt[n]{x_i}}\Bigg)\Bigg(1+\prod_{i=1}^{n}\sqrt[n]{x_i}\Bigg)$$

If $a_i=0$ for some $i$ then obvious Let $sin a_i=\frac{1}{x_i^n},x_i\geq 1$ then we get $$\Bigg(\frac{1}{n}\sum_{i=1}^{n}\frac{x_i^n}{1+x_i^n}\Bigg)\Bigg(1+\frac{1}{\prod_{i=1}^{n}x_i}\Bigg)\leq1$$or $$\sum_{i=1}^{n}\frac{1}{1+x_i^n} \geq \frac{n}{1+\prod_{i=1}^{n}x_i}$$which is IMO ShortList 1998, algebra problem 2

JANMATH111 wrote: Do you think I will get any points for that? I don't think so, you lost almost every equality case.

RagvaloD wrote: If $a_i=0$ for some $i$ then obvious Let $sin a_i=\frac{1}{x_i^n},x_i\geq 1$ then we get $$\Bigg(\frac{1}{n}\sum_{i=1}^{n}\frac{x_i^n}{1+x_i^n}\Bigg)\Bigg(1+\frac{1}{\prod_{i=1}^{n}x_i}\Bigg)\leq1$$or $$\sum_{i=1}^{n}\frac{1}{1+x_i^n} \geq \frac{n}{1+\prod_{i=1}^{n}x_i}$$which is IMO ShortList 1998, algebra problem 2 Very nice sir.