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Math Help - Let x and y be positive real numbers...

  1. #1
    MHF Contributor alexmahone's Avatar
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    Let x and y be positive real numbers...

    Let x and y be positive real numbers such that y^3 + y \le x - x^3. Prove that
    (a) y < x < 1; and
    (b) x^2 + y^2 < 1.
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  2. #2
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    From the given, x > 0\,\& \,y > 0, we get
    \begin{array}{rcl}   {y^3  + y \leqslant x - x^3 } &  \Leftrightarrow  & {y\left( {y^2  + 1} \right) \leqslant x\left( {1 - x^2 } \right)}  \\   {} &  \Leftrightarrow  & {0 < \frac{y}{x} \leqslant \frac{{1 - x^2 }}{{y^2  + 1}}}  \\   {} &  \Leftrightarrow  & {0 < 1 - x^2 }  \\   {} &  \Leftrightarrow  & {0 < x < 1}  \\<br />
 \end{array} .

    From the above we get
    y \geqslant x \Rightarrow \quad y^3  + y \geqslant x + x^3  \Rightarrow \quad y^3  + y > x - x^3 .

    Do you see ho to finish?
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  3. #3
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    a) y^3 + y is positive because y>0, hence x - x^3 > 0 \Rightarrow x ( 1 - x^2 ) > 0 \Rightarrow 1 - x^2 > 0 \Rightarrow x < 1
    y^3 + y \le x - x^3 <br />
\Rightarrow y^3 + x^3  \le x - y
    Because LHS is greater than 0 ( x and y are positive real numbers ) hence
    x-y > 0 and x>y

    b) x,y< 1
    Hence x^2 > x^4 (1)and
    y^2 > y^4 (2)

    Take away (2) from (1) and factorise \Rightarrow x^2 - y^2 > ( x^2 - y^2 ) ( x^2 + y^2 )

    Now x > y hence x^2 - y^2 > 0 and we can divide both side by x^2 - y^2 > 0 and get the desired inequality
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