# Hard calc problem

1. Sep 17, 2006

### ashrafmod

prove if a+b+c=1 ,a,b,c>0
so (1+1\a)(1+1\b)91+1\c)>=64

2. Sep 17, 2006

### ashrafmod

(1+1\a)(1+1\b)(1+1\c) >= 64

3. Sep 18, 2006

### dextercioby

Here's the solution

$$\frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}= \frac{abc+ab+ac+bc+2}{abc} =\frac{2}{abc}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1$$.

Now, for 3 arbitrary positive real numbers the harmonic average is smaller or equal to the arithmetic average

$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \leq \frac{a+b+c}{3}=\frac{1}{3}$$

from which it follows that

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 9$$

For 3 arbitrary positive numbers, the geometric average is smaller or equal to the arithmetic average

$$\sqrt[3]{abc} \leq \frac{a+b+c}{3}=\frac{1}{3}$$

from which it follows that

$$\frac{1}{abc} \geq 27$$.

Now i think you easily get the wanted inequality.

Daniel.

4. Sep 19, 2006

### jpr0

Dexterciboy, your signature made me chuckle. It reminded me of a joke we have about one of the professors in our department, who's supposed motto we claim to be:

"Never let an experiment get in the way of a good theory!"