Two exercises about real numbers

Hello!!

Could you please help me with these two exercises

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

Prove that: $\displaystyle \displaystyle\frac{a}{(a+1)(b+1)} + \displaystyle\frac{b}{(b+1)(c+1)} + \displaystyle\frac{c}{(a+1)(c+1)} \geq{\displaystyle\frac{3}{4}}$

Hint: $\displaystyle \sqrt[ ]{xy} \leq{\displaystyle\frac{x+y}{2}}$

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

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

Thanks a lot!!

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.

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

similar for b and c.

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

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

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

I'm a bit rushed just now.

Ask if you need more hints.

Re: Two exercises about real numbers

Re: Two exercises about real numbers

$\displaystyle a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\ge 6$

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

Now, using the fact that abc=1..

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

Now factorise the left hand side,

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

and then rearrange a little,

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

Nearly there...