ex(ex− 1) (1 + ex)3 which implies that f is convex in [0

Problem 11843

(American Mathematical Monthly, Vol.122, May 2015) Proposed by M. Bencze (Romania).

Let n and k be positive integers, and let xj ≥ 1 for 1 ≤ j ≤ n. Let y =Qn

i=1xi. Show that

n

X

i=1

1 1 + xi

≥

n

X

j=1

1

1 + (x^k−1_j y)^1/(n+k−1).

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

Let f (x) = 1/(1 + e^x) then

f⁰⁰(x) = e^x(e^x− 1) (1 + e^x)³

which implies that f is convex in [0, +∞). Since ln(x_j) ≥ 0 for 1 ≤ j ≤ n, it follows that Pn

i=1f (ln(xi)) + (k − 1)f (ln(xj))

n + k − 1 ≥ f

 Pn

i=1ln(xi) + (k − 1) ln(xj) n + k − 1

 , that is

1 n + k − 1

n

X

i=1

1

1 + x_i + k − 1 1 + x_j

!

≥ 1

1 + (x^k−1_j y)^1/(n+k−1). Hence, by summing over j, we obtain

1 n + k − 1

n

X

j=1 n

X

i=1

1 1 + xi

+ k − 1 1 + xj

!

≥

n

X

j=1

1

1 + (x^k−1_j y)^1/(n+k−1), which yields the required inequality as soon as we note that

n

X

j=1 n

X

i=1

1 1 + xi

+ k − 1 1 + xj

!

= n

n

X

i=1

1 1 + xi

+ (k − 1)

n

X

j=1

1 1 + xj

= (n + k − 1)

n

X

i=1

1 1 + xi

.