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Let $a,b,c,d,e,f$ be non negative real numbers. Find the minimum and maximum value of $ab+bc+cd+de+ef+fa$ given that $a+b+c+d+e+f=12$.

The minimum value will come as $0$ since we can take $a=b=c=d=e=0$ and $f=12$.

But how to find the maximum value? I tried to rearrange the terms but I couldn't be successful. I feel that the maximum value is $36$ since we can put $a=b=6$ and set the rest of the terms as $0$ and then the maximum value becomes $36$.

But these are just experimental verifications but can someone provide a strong, rigorous proof for both minimum and maximum value? Why can't we apply $AM-GM$ to get the minimum value?

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  • $\begingroup$ what Kind of numbers are $a,b,c,d,e,f$? $\endgroup$
    – Dr. Sonnhard Graubner
    May 25, 2017 at 8:32

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For $a=b=c=d=e=0$ and $f=12$ we get a value $0$, which is a minimal value.

Since $(a+b+c+d+e+f)^2\geq4(ab+bc+cd+de+ef+fa)$, which is $$(a-b+c-d+e-f)^2+A\geq0,$$ where $A\geq0$, we get a maximal value: $36$.

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  • $\begingroup$ Could you justify the last inequality? The asker may not understand immediately why. $\endgroup$
    – Parcly Taxel
    May 25, 2017 at 8:42
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    $\begingroup$ @Parcly Taxel I fixed my post. See now, please. $\endgroup$
    – Michael Rozenberg
    May 25, 2017 at 8:44

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