Find all functions from reals to reals satisfying
f(x+y) + f(y+z) + f(z+x) ≥ 3f(x+2y+3z)
for all x, y, z Belonging to reals.
-
UP 0 DOWN 0 0 1
1 Answers
341Nice. f(x) = c for some real c≥0 is the only family of functions satisfying the given relation.
Proof: Putting y=z=0, we get 2 f(x) + f(0) ≥ 3 f(x) or f(0)≥f(x) for all x
Again putting x=y=-z, we get f(-2z)≥f(0) for all z in other words f(t)≥f(0) for all real t.
This means f(x) = f(0) for all real x i.e. f(x) is a constant
