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Switzerland Contests
Switzerland - Final Round
2007 Switzerland - Final Round
5
5
Part of
2007 Switzerland - Final Round
Problems
(1)
f(xf(y)) f(y) = f (xy/(x + y) )
Source: Switzerland - 2007 Swiss MO Final Round p5
12/26/2022
Determine all functions
f
:
R
≥
0
→
R
≥
0
f : R_{\ge 0} \to R_{\ge 0}
f
:
R
≥
0
→
R
≥
0
with the following properties: (a)
f
(
1
)
=
0
f(1) = 0
f
(
1
)
=
0
, (b)
f
(
x
)
>
0
f(x) > 0
f
(
x
)
>
0
for all
x
>
1
x > 1
x
>
1
, (c) For all
x
,
y
≥
0
x, y\ge 0
x
,
y
≥
0
with
x
+
y
>
0
x + y > 0
x
+
y
>
0
holds
f
(
x
f
(
y
)
)
f
(
y
)
=
f
(
x
y
x
+
y
)
f(xf(y))f(y) = f\left( \frac{xy}{x + y}\right)
f
(
x
f
(
y
))
f
(
y
)
=
f
(
x
+
y
x
y
)
functional
functional equation
algebra