<s>[INST] Prove that if $0 < a,b,c <1$, then $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$. 
Prove that if $0 < a,b,c <1$, then $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1$. 

I think that using AM-GM might work. Thus we have $\dfrac{a+b+c}{3}+\sqrt{(1-a)(1-b)(1-c)} \geq \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}$ but I am not sure how to proceed. Maybe the maximum occurs when $a=b=c$ but how to prove that?
 [/INST] Note that $c<1$ and $1-c<1$ so :
$$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt{ab}+\sqrt{(1-a)(1-b)}$$
Now use AM-GM :
$$\sqrt{ab}+\sqrt{(1-a)(1-b)} \leq \frac{a+b}{2}+\frac{(1-a)+(1-b)}{2} =1$$ and so the conclusion follows .
</s>