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Math Help - Inequality

  1. #1
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    Inequality

    if p > q > r > 0

    prove p^4 + q^4 + r^4 > p^2 q^2 +q^2 r^2 +r^2 p^2
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  2. #2
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    p > q > 0\; \Rightarrow \;p^2  > pq\; \Rightarrow \;p^4  > p^2 q^2
    Can you use this to complete the problem?
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  3. #3
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    Quote Originally Posted by Plato View Post
    p > q > 0\; \Rightarrow \;p^2  > pq\; \Rightarrow \;p^4  > p^2 q^2
    Can you use this to complete the problem?
    I get up to p^4 + q^4 > p^2 q^2 +q^2 r^2<br />

    but then i keep going around in circles.
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  4. #4
    Senior Member JaneBennet's Avatar
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    Use the Cauchy–Schwarz inequality:

    p^2q^2+q^2r^2+r^2p^2\ \leq\ \left[(p^2)^2+(q^2)^2+(r^2)^2\right]^{\frac{1}{2}}\left[(q^2)^2+(r^2)^2+(p^2)^2\right]^{\frac{1}{2}}

    Equality obtains if and only if the vectors \left(\begin{array}{c}p^2\\q^2\\r^2\end{array}\rig  ht) and \left(\begin{array}{c}q^2\\r^2\\p^2\end{array}\rig  ht) in \mathbb{R}^3 are linearly dependent; i.e. if and only if p^2=q^2=r^2. Since p>q>r>0, this won’t happen and so the inequality is always strict.
    Last edited by JaneBennet; December 29th 2007 at 08:00 AM.
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  5. #5
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    Quote Originally Posted by bobak View Post
    if p > q > r > 0

    prove p^4 + q^4 + r^4 > p^2 q^2 +q^2 r^2 +r^2 p^2
    a^2 + b^2 + c^2 \geq ab+bc+ca
    (By AM-GM, \Sigma_{a,b,c}(a^2 + b^2) \geq \Sigma_{a,b,c}2ab)

    So substitute a=p^2,b=q^2,c=r^2

    The above AM-GM holds iff a=b=c.
    Or if p^2=q^2=r^2

    But since p>q>r, this cannot happen. So the inequality is strict. By the way, they need not be positive for AM-GM application here(since we are applying it to squares of those numbers)
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  6. #6
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Isomorphism View Post
    By the way, they need not be positive for AM-GM application here(since we are applying it to squares of those numbers)
    They need to be positive for the inequality to be strict. For example, if p=1,\ q=-1,\ r=1, the inequality will not be strict.
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  7. #7
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    thanks Isomorphism, i am familiar with the AM-GM inequality but i was hoping this could be pulled out by fiddling.

    JaneBennet the Cauchy–Schwarz inequality is totally alien to me, you know any good article with an explanation / proof of it ?
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  8. #8
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    Quote Originally Posted by JaneBennet View Post
    They need to be positive for the inequality to be strict. For example, if p=1,\ q=-1,\ r=1, the inequality will not be strict.
    q>r does not hold for your example.

    That is p>q>r is enough, r>0 need not hold. Thats what I was trying to say
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  9. #9
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    Quote Originally Posted by Isomorphism View Post
    a^2 + b^2 + c^2 \geq ab+bc+ca
    (By AM-GM, \Sigma_{a,b,c}(a^2 + b^2) \geq \Sigma_{a,b,c}2ab)
    This is C-S not AM-GM.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    This is C-S not AM-GM.
    No, I used AM-GM to get the result.Perhaps I should be more clear.

    By AM-GM,

    (a^2 + b^2) \geq 2ab
    (b^2 + c^2) \geq 2bc
    (c^2 + a^2) \geq 2ca

    Add them up,
    2(a^2 + b^2 + c^2) \geq 2ab + 2bc + 2ca

    Cancel 2 on both sides to wrap it up
    Yes I am aware that you can get it by a simple application of CS. But when we were in school we were taught only AM-GM. So I thought probably the poster wanted the proof using that
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  11. #11
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    Or you can just use C-S:

    (a^2+b^2+c^2) \geq \sqrt{ab+bc+ac}\cdot \sqrt{ab+bc+ac}
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