This is problem 2002 from *Mathematics Magazine*, v 82, n 4 (2009), p 313. The problem has been posted by Dorin Marghidanu and the solution by Northwestern University Math Problem Solving Group.

Problem

Let a_1, a_2, \dots, a_n be positive real numbers. Then

\frac{a_{1}^2}{{a_1}+{a_2}} + \frac{a_{2}^2}{{a_2}+{a_3}} + \dots + \frac{a_{n-1}^2}{{a_{n-1}}+{a_n}} + \frac{a_{n}^2}{{a_n}+{a_1}} \ge \frac{1}{2}\sum_{i=1}^{n} {a_i}.

Solution

Let a_{n+1} = a_1 and introduce

S = \sum_{i=1}^{n} \frac{a_{i}^2}{{a_i}+{a_{i+1}}}, T = \sum_{i=1}^{n} \frac{a_{i+1}^2}{{a_i}+{a_{i+1}}}.

Then

S - T = \sum_{i=1}^{n} \frac{{a_{i}^2 - a_{i+1}^2}}{{a_i}+{a_{i+1}}}

so that

S - T = \sum_{i=1}^{n} (a_{i+1} - a_{i}) = 0.

T = S. Now, for all positive x and y we have an easily proved *quadratic mean - arithmetic mean inequality*

\frac {x^2 + y^2}{x + y} \ge \frac{x + y}{2}.

Applying it to S + T gives

S + T = \sum_{i=1}^{n} \frac{a_{i}^2 + a_{i+1}^2}{a_{i} + a_{i+1}}

\ge \frac{1}{2} {\sum (a_i + a_{i+1}) = \sum {a_i}}.

\ge \frac{1}{2} {\sum (a_i + a_{i+1}) = \sum {a_i}}.

Since S = T this implies the required inequality.

Alternative Solution (Stuart Mark Anderson)

Let

S = \sum_{i=1}^{n} \frac{a_{i}^2}{{a_i}+{a_{i+1}}}

as above, and reindex this cyclic sum by i \rightarrow n-i. Writing out the terms in reverse order reveals that this is T. Hence S=T, and the rest of the solution proceeds as above.