Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$, a_1+a_2+...+a_n=n$. Is this true?

$$a_1a_2+a_2a_3+...+a_na_1\leq n$$

By observing:

When $n=1$, this is trivial;

When $n=2$, $ab\leq(\frac {a+b} 2)^2=1\leq2$;

When $n=3$, $ab+bc+ca\leq(\frac {a+b} 2)^2+(\frac {b+c} 2)^2+(\frac {c+a} 2)^2=\frac 1 2(a+b+c)^2-\frac 1 2(ab+bc+ca)$

$\Rightarrow ab+bc+ca\leq3$;

When $n=4$, $ab+bc+cd+da=(a+c)(b+d)\leq(\frac{a+b+c+d} 2)^2=4$.

But I can't find a more general way to prove these at once. Please help.

Thank you.

  • $\begingroup$ No I do not think so. Because if you try it by induction, it is not possible. If I write my method, it would look too unclear, but you better try it yourself by induction on paper. $\endgroup$ – Rohinb97 Jun 3 '13 at 11:33

The inequality does not hold in general. Indeed, take $a_1=a_2=\frac{n}{2}-1$ and $a_3=..=a_n=\frac{2}{n-2}.$ Then the left hand side of our inequality is greater that $n^2/4-n+1$ which can be made greater than $n.$

  • $\begingroup$ It should be $a_3=..=a_n=\frac{2}{n-2}$ to satisfy $\sum a_i = n$. $\endgroup$ – Milind Jun 3 '13 at 10:40
  • $\begingroup$ Explicitly, $n=8$ gives $3,3,\frac13,\frac13,\frac13,\frac13,\frac13,\frac13$. LHS$= 11+\frac59>8$ $\endgroup$ – ronno Jun 3 '13 at 12:47

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