Question

Fie x,y €(0,+infinit) si a>0 a diferit de 1. Sa se demonstreze ca
a)
$log_{a}( \frac{x + y}{2} ) \geqslant \frac{ log_{a}(x) + log_{a}(y) }{2} < = > a > 1$
b)
$log_{a}( \frac{x + y}{ 2} ) \leqslant \frac{ log_{a}(x) + log_{a}(y) }{2} < = > 0 < a < 1$
​

Answer 1

Răspuns:

Explicație pas cu pas:

a) Daca a > 1 si b > c, atunci log(a;b) > log(a;c)

log(a;(x+y)/2) >= log(a;xy)/2

2*log(a;(x+y)/2) >= log(a;xy)

log(a;(1/4*(x+y)^2) >= log(a;xy)

1/4*(x+y)^2 >= xy

x^2 +2xy +y^2 >= 4xy

x^2 -2xy +y^2 >= 0

(x-y)^2 >= 0 evident

b) Daca 0 < a < 1  si b < c, atunci log(a;b) > log(a;c)

log(a;(x+y)/2) <= log(a;xy)/2

2*log(a;(x+y)/2) <= log(a;xy)

log(a;1/4*(x+y)^2) <= log(a;xy)

1/4*(x+y)^2 >= xy

Se ajunge la acelasi rezultat