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Math Help - Urgent Inequality Help Please

  1. #1
    Junior Member
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    Thumbs up Urgent Inequality Help Please

    Prove that the inequality
    3a4 - 4a3b + b4 ≥ 0
    holds for all real numbers a and b.


    THANK YOU VERY MUCH FOR YOUR HELP!!!
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  2. #2
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    Hello, rlarach!

    Prove that: . 3a^4 - 4a^3b + b^4 \;\geq \;0 for all real numbers a and b

    Let f \:=\:3a^4-4a^3b + b^4

    I used the Factor Theorem a few times.

    If b = a, the polynomial has value 0.
    . . Hence, (a -b) is a factor of f.

    Using long division: . 3a^3 -4a^3b + b^4\;=\;(a - b)(3a^3 - a^2b - ab^2 - b^3)

    And we find that (a-b) is a factor of the cubic.

    Long division: . f \;=\;(a-b)(a-b)(3a^2+2ab+b^2)

    . . Hence, we have: . f \;=\;(a-b)^2\left(2a^2 + [a+b]^2\right)


    We see that: (a-b)^2 \:\geq\:0

    . . and that: . 2a^2 + (a+b)^2\:\geq\;0


    Therefore, f \:\geq\:0 for all real a,\,b

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