# Two exercises about real numbers

• Apr 16th 2012, 12:50 PM
osodud
Hello!!

1.- Let $a,b,c$ be real positive numbers and $abc = 1$

Prove that: $\displaystyle\frac{a}{(a+1)(b+1)} + \displaystyle\frac{b}{(b+1)(c+1)} + \displaystyle\frac{c}{(a+1)(c+1)} \geq{\displaystyle\frac{3}{4}}$
Hint: $\sqrt[ ]{xy} \leq{\displaystyle\frac{x+y}{2}}$

2.- Let $a,b,c$ be real positive numbers and $a+b+c = 1$

Prove that: $(\displaystyle\frac{1}{b} + 1)$ $(\displaystyle\frac{1}{b} + 1)$ $(\displaystyle\frac{1}{c} + 1)$ $\geq{64}$

Thanks a lot!!
• Apr 17th 2012, 02:19 AM
a tutor
Re: Two exercises about real numbers
Hi. I worked from the inequality towards something I know is true and then rewrote the proof in the correct direction.

It's a bit clumsy but here's how I started.

$(a-1)^2\ge 0$

similar for b and c.

$(a-1)^2\ge 0 \Rightarrow a+\frac{1}{a}\ge 2$

add three of those together to get $a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\ge 6$

Now add 2 to both sides and ask yourself what is another way to write $\frac{1}{a}$ etc and what is another way to write 1?

I'm a bit rushed just now.

Ask if you need more hints.
• Apr 17th 2012, 10:02 PM
osodud
Re: Two exercises about real numbers
• Apr 17th 2012, 11:39 PM
a tutor
Re: Two exercises about real numbers
$a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\ge 6$

$a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}+2\ge 8$

Now, using the fact that abc=1..

$a+bc+b+ac+c+ab+abc+1\ge 8$

Now factorise the left hand side,

$(a+1)(b+1)(c+1)\ge8$

and then rearrange a little,

$1-\frac{2}{(a+1)(b+1)(c+1)}\ge\frac{3}{4}$.

Nearly there...