Apply Cauchy_Schwarz inequality and find that the LHS is at most \sqrt bc+(1-b)(1-c) which is smaller than 1.

In fact i was going to post this here too. i didn't understand your solution harazi. Any help please?

Nickolas, LaTeX is really easy to learn, at least the few basics needed to formulate a problem like yours: If a, b, c are real numbers in the open interval ]0; 1[ (some people also denote this interval by (0; 1)), then prove the inequality $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$. Harazi, I admit I don't understand your solution, but here is mine: I will use the following inequality: Let k and n be natural numbers, and $a_{i;j}$ some positive reals for all natural i and j with $1\leq i\leq k$ and $1\leq j\leq n$. Moreover, let $p_i$ be positive reals for all natural i with $1\leq i\leq k$. We assume that $p_{1}+p_{2}+...+p_{k}=1$. Then, $a_{1;1}^{p_{1}}a_{2;1}^{p_{2}}...a_{k;1}^{p_{k}}+a_{1;2}^{p_{1}}a_{2;2}^{p_{2}}...a_{k;2}^{p_{k}}+...+a_{1;n}^{p_{1}}a_{2;n}^{p_{2}}...a_{k;n}^{p_{k}}$ $\leq \left( a_{1;1}+a_{1;2}+...+a_{1;n}\right) ^{p_{1}}\cdot \left( a_{2;1}+a_{2;2}+...+a_{2;n}\right) ^{p_{2}}\cdot ...\cdot \left( a_{k;1}+a_{k;2}+...+a_{k;n}\right) ^{p_{k}}$. This was proven in Problem 304 of D. O. Shkljarskij, N. N. Chenzov, I. M. Yaglom, Izbrannye Zadachi i Teoremy Elementarnoj Matematiki, Chastj 1 (Arifmetika i Algebra), Moscow 1954 (I believe this book was translated into English and published as the "USSR olympiad problem book" or something like this). It is actually one form of (one possible?) generalization of the Hölder inequality. Now apply it to k = 3, n = 2, $a_{1;1}=a$, $a_{2;1}=b$, $a_{3;1}=c$, $a_{1;2}=1-a$, $a_{2;2}=1-b$, $a_{3;2}=1-c$ (note that all of these variables are positive), and $p_{1}=p_{2}=p_{3}=\frac{1}{3}$. Then you get $\root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }$ $\leq \root{3}\of{a+\left( 1-a\right) }\cdot \root{3}\of{b+\left( 1-b\right) }+\root{3}\of{c+\left( 1-c\right) }=1$. Now the inequality $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$ trivially follows, since abc < 1 and (1 - a) (1 - b) (1 - c) < 1 imply $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<\root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left(1-c\right) }$. Darij

I think harazi meant $ \sqrt {a} \sqrt{bc} + \sqrt {1-a} \sqrt{(1-b)(1-c)} \leq \sqrt{a+1-a}\sqrt{bc + (1-b)(1-c)} = \sqrt{1-b-c} < 1 $

Yes.

billzhao wrote: I think harazi meant $ \sqrt {a} \sqrt{bc} + \sqrt {1-a} \sqrt{(1-b)(1-c)} \leq \sqrt{a+1-a}\sqrt{bc + (1-b)(1-c)} = \sqrt{1-b-c} < 1 $ Thanks. But you have a little mistake in your last calculation: Actually, bc + (1 - b) (1 - c) = 1 - (b + c) + 2bc. Therefore, the corrected proof should be $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }=\sqrt{a}\sqrt{bc}+\sqrt{1-a}\sqrt{\left( 1-b\right) \left( 1-c\right) }$ $\leq \sqrt{a+\left( 1-a\right) }\sqrt{bc+\left( 1-b\right) \left( 1-c\right) }=\sqrt{bc+\left( 1-b\right) \left( 1-c\right) }=\sqrt{1-\left( b+c\right) +2bc}<1$. Hereby, $\sqrt{1-\left( b+c\right) +2bc}<1$ is because $1-\left( b+c\right) +2bc<1$, what itself follows from $2bc\leq b^{2}+c^{2}<b+c$ (in fact $b^2 < b$ and $c^2 < c$ since b < 1 and c < 1). Thanks again, I understand now. Darij

darij grinberg wrote: $\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<\root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left(1-c\right) }$. Darij After noting this, we may use AM-GM here to finish the proof. $\displaystyle \root{3}\of{abc}+\root{3}\of{\left( 1-a\right) \left( 1-b\right) \left(1-c\right) } \leq \frac{a+b+c}{3} + \frac{(1-a)+(1-b)+(1-c)}{3} = 1$.

If $a, b, c$ are real numbers in the open interval $ (0,1)$ then prove the inequality

$$\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$$PS

parmenides51 wrote: If $a, b, c$ are real numbers in the open interval $ (0,1)$ then prove the inequality

$$\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }<1$$PS

Junior Balkan Team Selection Tests - Romania 2002 See also here $$\sqrt{abc}+\sqrt{\left( 1-a\right) \left( 1-b\right) \left( 1-c\right) }< \sqrt{ab}+\sqrt{(1-a)(1-b)}\leq \frac{a+b}{2}+\frac{1-a+1-b}{2}=1$$