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Thread: Abstract Inequalities

  1. #1
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    Abstract Inequalities

    Q. Having some trouble with this one. Can anyone help me out?


    Many thanks.


    Q. Use the result, $\displaystyle a^2+b^2\geq2ab$, to prove that $\displaystyle a^2+b^2+c^2\geq ab+bc+ca$ for all real values of a, b & c.


    Attempt: $\displaystyle \frac{a^2+b^2}{2}\geq ab$
    if $\displaystyle \frac{a^2+b^2}{2}-ab\geq0$
    if $\displaystyle \frac{(a+b)(a-b)}{2}-ab\geq0$
    if $\displaystyle \frac{a^2-ab+ab+b^2}{2}-ab\geq0$
    if$\displaystyle \frac{a^2-ab+ab+b^2-2ab}{2}\geq0$
    if $\displaystyle \frac{a^2-2ab+b^2}{2}$
    if $\displaystyle \frac{(a-b)^2}{2}\geq0$


    Thus: $\displaystyle a^2+b^2+c^2\geq ab+bc+ca$
    if $\displaystyle \frac{2a^2+2b^2+2c^2}{2}-ab-bc-ca\geq0$
    if $\displaystyle \frac{2a^2+2b^2+2c^2-2ab-2bc-2ca}{2}\geq0$
    if $\displaystyle a^2-b^2-c^2-ab-bc-ca\geq0$
    if $\displaystyle (a+b+c)^2\geq0$
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  2. #2
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    Re: Abstract Inequalities

    Quote Originally Posted by GrigOrig99 View Post
    Q.
    Q. Use the result, $\displaystyle a^2+b^2\geq2ab$, to prove that $\displaystyle a^2+b^2+c^2\geq ab+bc+ca$ for all real values of a, b & c.
    $\displaystyle \begin{align*} a^2+b^2 &\ge 2ab \\ a^2+c^2 &\ge 2ac \\ b^2+c^2&\ge 2bc \\ \text{Add all three }2a^2+2b^2+2c^2 &\ge 2ab+2ac+2bc \end{align*}$
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  3. #3
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    Re: Abstract Inequalities

    Great, thank you very much.
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