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Thread: 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 $\displaystyle y^3 + y \le x - x^3$. Prove that
    (a) $\displaystyle y < x < 1$; and
    (b) $\displaystyle x^2 + y^2 < 1$.
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  2. #2
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    From the given, $\displaystyle x > 0\,\& \,y > 0$, we get
    $\displaystyle \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} \\
    \end{array} $.

    From the above we get
    $\displaystyle 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) $\displaystyle y^3 + y$ is positive because y>0, hence $\displaystyle x - x^3 > 0 \Rightarrow x ( 1 - x^2 ) > 0 \Rightarrow 1 - x^2 > 0 \Rightarrow x < 1$
    $\displaystyle y^3 + y \le x - x^3
    \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 $\displaystyle x^2 > x^4$ (1)and
    $\displaystyle y^2 > y^4$ (2)

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

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