(1)Problem 12149 (American Mathematical Monthly, Vol.126, December 2019) Proposed by M

Problem 12149

(American Mathematical Monthly, Vol.126, December 2019) Proposed by M. Mehrabi (Sweden).

Prove

x^xy^y



Γ x + y 2

²

≤ x + y 2

^x+y

Γ(x) Γ(y) for all positive real numbers x and y.

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

Solution. Let

f(x) = log Γ(x) x^x



then the inequality is equivalent to

f x + y 2



≤ f(x) + f (y)

2 .

Hence it suffices to show that the second derivative of f is positive which implies that f is convex in (0, +∞). Recall that

Γ(x) = lim

N →∞

N^xN!

x(x + 1) (x + 2) · · · (x + N ) and it follows that

(log(Γ(x)))^′′= lim

N →∞ xlog(N ) + log(N !) −

N

X

n=0

log(x + n)

!^′′

=

∞

X

n=0

1 (x + n)². Therefore

f^′′(x) = (log(Γ(x)))^′′−(x log(x))^′′=

∞

X

n=0

1

(x + n)² −1 x

>

∞

X

n=0

1

(x + n)(x + n + 1)−1 x =

∞

X

n=0

 1

x+ n − 1 x+ n + 1



−1 x= 1

x− 1 x= 0.