Prove this inequality $(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n}$ Assmue that $a_{i}\in (0,1),i=1,2,3,\cdots,n$,show that
$$(a_{1}a_{2}\cdots a_{n})\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}<\sqrt{n},$$
I've tried many things but all have failed.

If $u\in(0,1)$ then, by Cauchy-Schwarz,
$$ \sqrt{\frac1n}\,u + \sqrt{1-\frac1n}\,(1-u^2)
\le \sqrt{u^2 + (1-u^2)^2} < \sqrt{u^2 + (1 - u^2)} = 1 $$
Take $u=\sqrt{1-a_{n+1}}$ and rearrange to get
$$ \sqrt{1-a_{n+1}} + \sqrt{n-1}\,a_{n+1} < \sqrt n $$
Introducing $a_1\dotsm a_n$ on the first term on the left only makes the left smaller.
