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    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #1

    Sep 23, 2010, 08:29 AM
    inequality
    if a,b ∈ R >>> sqrt(a^2 + b^2)>= (|a|+|b|)/sqrt(2)

    sqrt(a^2+(1-a^2))+sqrt(b^2+(1-b^2))+sqrt(c^2+(1-c^2))>=(3sqrt(2))/2??

    please give some hints. I have no idea how to solve it.
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #2

    Sep 23, 2010, 09:07 AM
    inequality
    if a,b>0 >>> (a^2+b^2)/(a+b)>=(a+b)/2
    if a=b=0 >>> (a^2+b^2)/(a+b)=(a+b)/2

    (a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=ab+bc+ca, for any a,b,c>0
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
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    #3

    Sep 23, 2010, 09:23 AM
    is it correct if I say:
    (a^4+b^4)/(a^2+b^2)>=(a^2+b^2)/2
    (b^4+c^4)/(b^2+c^2)>=(b^2+c^2)/2
    (c^4+a^4)/(c^2+a^2)>=(c^2+a^2)/2
    ____________________________
    (a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=[2(a^2+b^2+c^2)]/2
    (a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=(a^2+b^2+c^2)
    (a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=[(a+b+c)^2]-2(ab+bc+ca)
    (a^4+b^4)/(a^2+b^2)+(b^4+c^4)/(b^2+c^2)+(c^4+a^4)/(c^2+a^2)>=(a+b+c)^2>=2(ab+bc+ca)

    ??
    lemon14's Avatar
    lemon14 Posts: 143, Reputation: 9
    Junior Member
     
    #4

    Sep 23, 2010, 09:50 AM
    I think I got it:
    sqrt(a^2+(1-a^2))>=(|a|+|1-a|)/sqrt(2)
    sqrt(b^2+(1-b^2))>=(|b|+|1-b|)/sqrt(2)
    sqrt(c^2+(1-c^2))>=C

    (|a|+|1-a|)/sqrt(2)>=|a+1-a|/sqrt(2)=2/sqrt(2)
    >>>(|b|+|1-b|)/sqrt(2)>=2/sqrt(2) & (|b|+|1-b|)/sqrt(2)>=2/sqrt(2)
    >>>sqrt(a^2+(1-a^2))+sqrt(b^2+(1-b^2))+sqrt(c^2+(1-c^2))>=(3sqrt(2))/2

    let me know what you think

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    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Sep 23, 2010, 10:43 AM

    It would really help if you learned to use the LaTeX facility. And I have no idea of what you are trying to do.

    is it correct if i say:

    ____________________________


    ??
    The last line is not good. You cannot introduce this inequality sign like that.

    EDIT: See ebaines post. I didn't know that... well, it is plausible (the expression is still true/valid), but I didn't know we could just insert it like that... =/
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
    Expert
     
    #6

    Sep 23, 2010, 11:53 AM

    Actually I think the last line is good - if you known that m-n>0 (which you know from from the line above it), then m>n.

    I don't understand this one, from post #2:
    if a=b=0 >>> (a^2+b^2)/(a+b)=(a+b)/2

    If a= b = 0, then (a^2 + b^2)/(a+b) is undefined.

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