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National and Regional Contests
Moldova Contests
Moldova Team Selection Test
1999 Moldova Team Selection Test
3
$$f(x_1)+f(x_2)+\ldots+f(x_n)=nf(\sqrt[n]{x_1x_2\ldots x_n}),$$
$$f(x_1)+f(x_2)+\ldots+f(x_n)=nf(\sqrt[n]{x_1x_2\ldots x_n}),$$
Source: Moldova TST 1999
August 7, 2023
function
Problem Statement
The fuction
f
(
0
,
∞
)
→
R
f(0,\infty)\rightarrow\mathbb{R}
f
(
0
,
∞
)
→
R
verifies
f
(
x
)
+
f
(
y
)
=
2
f
(
x
y
)
,
∀
x
,
y
>
0
f(x)+f(y)=2f(\sqrt{xy}), \forall x,y>0
f
(
x
)
+
f
(
y
)
=
2
f
(
x
y
)
,
∀
x
,
y
>
0
. Show that for every positive integer
n
>
2
n>2
n
>
2
the following relation takes place
f
(
x
1
)
+
f
(
x
2
)
+
…
+
f
(
x
n
)
=
n
f
(
x
1
x
2
…
x
n
n
)
,
f(x_1)+f(x_2)+\ldots+f(x_n)=nf(\sqrt[n]{x_1x_2\ldots x_n}),
f
(
x
1
)
+
f
(
x
2
)
+
…
+
f
(
x
n
)
=
n
f
(
n
x
1
x
2
…
x
n
)
,
for every positive integers
x
1
,
x
2
,
…
,
x
n
x_1,x_2,\ldots,x_n
x
1
,
x
2
,
…
,
x
n
.
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