Show that n X i=1 (ai+ ai+1)4 a2i −aiai+1+ a2i+1 ≥12n, wherean+1= a1

Problem 11963

(American Mathematical Monthly, Vol.124, February 2017) Proposed by G. Alexe and G.F. Serban (Romania).

Leta1, . . . , an be positive real numbers with Qn

i=1ai= 1. Show that

n

X

i=1

(ai+ ai+1)⁴

a²_i −aiai+1+ a²_i+1 ≥12n, wherean+1= a1.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. We have that (x + y)⁴

x²−xy + y² −12xy = (x + y)⁴−12xy(x²−xy + y²)

x²−xy + y² = (x²−4xy + y²)² x²−xy + y² ≥0.

Hence

n

X

i=1

(ai+ ai+1)⁴

a²_i −aiai+1+ a²_i+1 ≥12

n

X

i=1

aiai+1≥12n

n

Y

i=1

aiai+1

!1/n

= 12n

n

Y

i=1

ai

!2/n

= 12n.

where in the second step we used the AM-GM inequality.

Note that 12 is the best constant because equality holds when a1=p3/2+p1/2, a2=p3/2−p1/2.