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FC Alan Walker

21/05/2018 at 02:51
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Prove that with all positive real numbers a, b, c, d, e we have:

                        \(\sum\dfrac{a+b}{2}\ge\sum\dfrac{a+b+c}{3}\)




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    FC Alan Walker 21/05/2018 at 02:54

    Sorry, \(\dfrac{a+b}{2}.\dfrac{b+c}{2}.\dfrac{c+d}{2}.\dfrac{d+e}{2}.\dfrac{e+a}{2}\ge\dfrac{a+b+c}{3}.\dfrac{b+c+d}{3}.\dfrac{c+d+e}{3}.\dfrac{d+e+a}{3}.\dfrac{e+a+b}{3}\)


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