(1)Problem 11713 (American Mathematical Monthly, Vol.120, June-July 2013) Proposed by Mihaly Bencze (Romania)

Problem 11713

(American Mathematical Monthly, Vol.120, June-July 2013) Proposed by Mihaly Bencze (Romania).

Let x1, . . . , xn be nonnegative real numbers. Prove that

n

Y

k=1

(1 + xk) ≤ 1 +

n

X

k=1

1 k!

 1 − k

2n

k−1 n

X

k=1

xk

!^k .

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

Let σk be k-th elementary symmetric polynomial of x1, . . . , xn,

σk= X

i1<i2<···<ik

xi₁xi₂· · · xi_k.

By the Symmetric Mean inequality, it follows that σk

n k

!1/k

≤ σ1

n. Hence

σk≤n k



·σ₁^k

n^k = (n − 1) · · · (n − (k − 1)) n^k−1 ·σ^k₁

k!

≤ (((n − 1) + · · · + (n − (k − 1)))/(k − 1))^k−1

n^k−1 ·σ₁^k

k! =

 1 − k

2n

^k−1

·σ^k₁ k!

where we used the AM-GM inequality. Finally,

n

Y

k=1

(1 + xk) = 1 +

n

X

k=1

σk ≤ 1 +

n

X

k=1

 1 − k

2n

k−1

·σ₁^k k!.