Find the maximum value of $ab+bc+cd+de+ef+fa,$ given that $a+b+c+d+e+f=1$

After looking at this post, I framed the following question.

If $$a,b,c,d,e,f$$ are positive real numbers such that $$a+b+c+d+e+f=1$$, then find the minimum and maximum value of $$ab+bc+cd+de+ef+fa$$

My Attempt:

$$\frac{a+b+c+d+e+f}6\ge(abcdef)^{\frac16}\implies\frac1{36}\ge(abcdef)^{\frac13}$$

Also, $$\frac{ab+bc+cd+de+ef+fa}6\ge(abcdef)^{\frac13}$$

Dividing the second inequality by first, I get,

$$ab+bc+cd+de+ef+fa\ge\frac16$$

I got the minimum value. For maximum value, I think this post has a solution, but not able to understand how it framed the inequalities.

Also, the minimum value could have been achieved by assigning each variable as $$\frac16$$. In such equality cases, how do we know whether we'll get minimum value or maximum value, without solving algebraically?

• your minimum is wrong. You are claiming that $A\geq B$ and $C\geq B$ implies $C\geq\max B.$ Or just set $a\to 1$ and $b,c,d,e,f\to 0$ and you will see that the minimum can be arbirtrarily small positive number. Feb 3 at 18:05
• @dezdichado thanks Feb 3 at 18:07
• I guess you can simply apply the approaches used in the linked posts to your problem. Note that: $$ab+bc+cd+de+ef+fa \\ =b(c+a)+d(c+e)+f(e+a)\\ \leq (b+d+f)(a+c+e)\\ \leq \frac {((b+d+f)+(a+c+e))^2}{4}=\frac{1}{4}.$$ Clearly, $\frac{1}{4}$ is attainable at $a=b=\frac{1}{2}$ and $c=d=e=f=0.$ Feb 3 at 19:05
• @RezaRajaei in the linked posts, the variables are non-negative, so, zero is allowed. I am taking variables as positives. Feb 3 at 19:27
• @RezaRajaei but yes, if we take variables to tend to zero then $\frac14$ seems like a good maximum value, thanks. Feb 3 at 19:30