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Math Help - Two exercises about real numbers

  1. #1
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    Two exercises about real numbers

    Hello!!
    Could you please help me with these two exercises

    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!!
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  2. #2
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    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.
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  3. #3
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    Re: Two exercises about real numbers

    Yes. more hints please.
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  4. #4
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    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...
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