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.

Let $C_n:=a_1\cdots a_n$. Note, that for $n=1$, the inequality is true:
$$
a_1\cdot\sqrt{1-a_{n+1}}<1\cdot\sqrt{1-0}=1
$$
Now we have the following for $n>1$:
$$
0≤\sqrt{n-1}\left(\sqrt{1-a_{n+1}}-\frac{C_n}{2\sqrt{n-1}}\right)^2 \iff \\
0≤\sqrt{n-1}\cdot (1-a_{n+1})-C_n\sqrt{1-a_{n+1}}+\frac{C_n^2}{4\sqrt{n-1}}\iff\\ C_n\sqrt{1-a_{n+1}}+\sqrt{n-1}\cdot a_{n+1}≤\sqrt{n-1}+\frac{C_n^2}{4\sqrt{n-1}}<\sqrt{n-1}+\frac{1}{4\sqrt{n-1}}
$$
Where in the last inequality we used $C_n<1$. Therefore, we have to prove 
$$
\sqrt{n-1}+\frac{1}{4\sqrt{n-1}}<\sqrt{n}\iff \frac{1}{4\sqrt{n-1}}<\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}\iff \\\sqrt{n}<3\sqrt{n-1}\iff n<9n-9 \iff \frac{9}{8}<n
$$
So we conclude that the original inequality is true.
