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Thread: inequality

  1. #1
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    inequality

    inequality-inequality1.jpg
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  2. #2
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    Re: inequality

    Quote Originally Posted by mulfer View Post
    Click image for larger version. 

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    Small advice @ mulferlease learn LATEX

    We have to prove:
    $\displaystyle a^2+b^2+c^2 \geq a+b+c$
    It is equivalent to proving:
    $\displaystyle a^2+b^2+c^2-a-b-c \geq 0$
    $\displaystyle \implies a(a-1)+b(b-1)c(c-1) \geq 0 $ So we are done if we are able to prove this.
    now,
    if $\displaystyle x$ is a positive real number,
    if$\displaystyle x>1$ then $\displaystyle x(x-1)>(x-1)$ (obvious)
    if $\displaystyle x=1 $then $\displaystyle x(x-1)=(x-1) $(obvious)
    if $\displaystyle 0<x<1$ then $\displaystyle (x-1)<0 $ so, $\displaystyle x<1$ $\displaystyle \implies x(x-1)>x-1$ (multiplying both sides by a negative number i.e $\displaystyle (x-1)$ changes the sign)
    so for all case, $\displaystyle x(x-1)\geq x-1$
    similarly,
    $\displaystyle a(a-1) \geq a-1$
    $\displaystyle b(b-1) \geq b-1$
    $\displaystyle c(c-1) \geq c-1$
    summing these we get:
    $\displaystyle a(a-1)+b(b-1)c(c-1) \geq (a-1)+(b-1)+(c-1) = (a+b+c)-3$
    so we will be done if we prove $\displaystyle (a+b+c)-3 \geq 0$ or $\displaystyle a+b+c \geq 3$
    by the AM-GM inequality,
    $\displaystyle (a+b+c)\geq 3(abc)^{1/3}=3$ # as $\displaystyle abc=1$
    proved
    Last edited by earthboy; Jan 27th 2013 at 05:30 AM.
    Thanks from mulfer
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  3. #3
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    Re: inequality

    thanks for replying. I'd be glad to learn the latex soon.
    please, check my solution:
    inequality-.jpg
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  4. #4
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    Re: inequality

    Quote Originally Posted by mulfer View Post
    thanks for replying. I'd be glad to learn the latex soon.
    please, check my solution:
    Click image for larger version. 

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    Yup..Its seems alright to me
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