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Math Help - Inequality

  1. #1
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    Inequality

    if x+y=2
    then show that (x^3+y^3)(x^3)(y^3)<=2
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  2. #2
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    Start by breaking x^{3}+y^{3} down.
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  3. #3
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    Thanks,got it..
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  4. #4
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    You could use the Arithmetic-Geometric mean inequality, for x,\;y\ \ge\ 0

    x+y=2

    x^3+y^3=(x+y)\left(x^2-xy+y^2\right)=2\left(x^2-xy+y^2\right)

    x^2-xy+y^2=(x+y)^2-3xy=4-3xy

    \left(x^3+y^3\right)x^3\;y^3=2(4-3xy)x^3\;y^3\ \le\ 2\;\;?

    (4-3xy)x^3\;y^3\ \le\ 1\;\;?

    AM-GM Inequality: \displaystyle\frac{x+y}{2}\ \ge\sqrt{xy}\Rightarrow\ xy\ \le\ 1

    (4-3xy)x^3\;y^3=\displaystyle\frac{4}{\left(\frac{1}{  x^3\;y^3}\right)}-\frac{3}{\left(\frac{1}{x^2\;y^2}\right)}

    The divisor of 4 is greater than than the divisor of 3.
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