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Math Help - Easy inequality

  1. #1
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    Easy inequality

    Let x,y be real. Show that if x^2 + y^2 > 2, then x^4 + y^4 > x^2 + y^2
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  2. #2
    Junior Member slovakiamaths's Avatar
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    Quote Originally Posted by EinStone View Post
    Let x,y be real. Show that if x^2 + y^2 > 2, then x^4 + y^4 > x^2 + y^2
    As x^2 + y^2 > 2 it means x and y are both greater than 1. so,
    x^4>x^2 ...............(i)
                              y^4>y^2................(ii)
    adding (i) & (ii), we have
    x^4 + y^4 > x^2 + y^2
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  3. #3
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    Quote Originally Posted by slovakiamaths View Post
    As \color{red}x^2 + y^2 > 2 it means x and y are both greater than 1. so,
    What makes you think that is true?
    It is not true: x=2~\&~y=0.5
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  4. #4
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    Quote Originally Posted by EinStone View Post
    Let x,y be real. Show that if x^2 + y^2 > 2, then x^4 + y^4 > x^2 + y^2
    If x = r \cos \theta and y = r \sin \theta, then you would like to establish that

    r^2 \left( \cos^4 \theta + \sin^4 \theta\right) > 1 \; \text{if}\; r^2 > 2 or

     <br />
\cos^4 \theta + \sin^4 \theta \ge \frac{1}{2}.<br />

    Since \cos^2\theta = 1 - \sin^2 \theta then we wish to prove that

     <br />
\sin^4 \theta +(1 - \sin^2 \theta)^2 \ge \frac{1}{2},<br />

    or

     <br />
2 \sin^4 \theta - 2\sin^2 \theta + 1 \ge \frac{1}{2}<br />

    Completing the square gives

     <br />
\left( \sin^2 \theta - \frac{1}{2}\right)^2 + \frac{1}{4} \ge \frac{1}{4}<br />

    which obviously is true.
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  5. #5
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    Assume, without loss of generality, that x^2\ge 1 \ge y^2.

    x^2+y^2 > 2 \Leftrightarrow x^2-1 > 1-y^2,

    where x^2-1 \ge 0 and 1-y^2 \le 0,

    so x^2(x^2-1)>y^2(1-y^2), and thus x^4+y^4 > x^2+y^2.
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