Consider the quantity b_n = a_(n+2) + 3a_(n+1) + a_n. It is easily verified that b_(n+2) - b_(n+1) <= b_(n+1) - b_n is a translation of the intial inequality condition. However, the b_i are positive since the a_i are positive. If d_n = b_(n+1) - b_n < 0, then b_(n+k) <= b_n +kd_n which is negative for sufficiently large k, a contradiction. We derive a similar contradiction in the other direction for d_n > 0. Therefore, d_n = 0 for all n and the b_i form a constant sequence. Suppose b_n = 5c for some constant c. Let a_n - c = c_n. The equation 5c = a_(n+2) + 3a_(n+1) + a_n becomes c_(n+2) + 3c_(n+1) + c_n = 0. This has characteristic equation t^2+3t+1=0. It follows that c_n has explicit formula c_n = A x^n + B x^(-n) where x=(-3-5^.5)/2. As n--> + infinity, x^n dominates x^(-n) and alternates sign implying A=0 (else a_n has negative terms). Similarly, n-->-infinity implies B=0. Therefore, c_n = 0 for all n and a_n = c as required. Note: The consideration of a_(n+2) + 3a_(n+1) + a_n may seem implausible. However, this comes from considering the equality case of the given inequality (where a_n=c will obviously occur). The recurrence has characteristic polynomial (t-1)^2 (t^2+3t+1) which leads naturally to the consideration of b_n.

Really, I can't understand why sequence given by the $a_{n}= \sqrt{n}$ is not a counter example? More generally, why every concave (and positive for natural numbers) function is not a counter example?

TomciO wrote: Really, I can't understand why sequence given by the $a_{n}= \sqrt{n}$ is not a counter example? More generally, why every concave (and positive for natural numbers) function is not a counter example? What is $a_{-1}$?

Ok, now I can see that it is mentioned that $n$ is an integer, thanks.