Problem:
Find all functions f:(0,∞)→(0,∞) such that
f(x+y1​)+f(y+z1​)+f(z+x1​)=1
for all x,y,z>0 with xyz=1.
Solution:
For any u,v,w∈(0,1) satisfying u+v+w=1, we may set x=vu​,y=wv​, and z=uw​ to obtain
f(wu+v​)+f(uv+w​)+f(vw+u​)=1
and thus
f(w1​−1)+f(u1​−1)+f(v1​−1)=1
First, let g:(0,1)→(0,∞) be given by g(x)=f(x1​−1), so that the above equation reads
g(u)+g(v)+g(w)=1 for all u,v,w∈(0,1) with u+v+w=1
Note that this condition implies actually g(x)<1 for all x.
Next, consider the function h:(−1/3,2/3)→(−1/3,2/3) given by h(x)=g(x+1/3)−1/3. Then, we have for all x,y,z∈(−1/3,2/3) with x+y+z=0 that
h(x)+h(y)+h(z)=0(1)
We now establish the key properties of h in a series of claims.
Claim 1. We have h(0)=0 and for all x∈(−1/3,1/3), we have h(−x)=−h(x).
Proof. Setting x=y=z=0 in (1) gives h(0)=0. Then, setting z=0 and y=−x yields h(−x)=−h(x), as long as x∈(−1/3,1/3).
Claim 2. For all x,y∈(0,2/3) with x+y<2/3, we have h(x+y)=h(x)+h(y).
Proof. In the case where x,y<1/3, we immediately have from Claim 1 and (1) that
h(x)+h(y)=−h(−x)−h(−y)=h(x+y)
This allows us to deduce the same property for all x and y satisfying the specified conditions. Indeed, we have
h(x+y)=h(2x+y​)+h(2x+y​)=2h(2x​)+2h(2y​)=h(x)+h(y)
where we have used the fact that x+y<2/3 implies x/2,y/2,(x+y)/2 are all less than 1/3.
Claim 3. For all x∈(−1/3,2/3), we have h(x)=3h(1/3)x.
Proof. Note that by repeated applications of Claim 2, we have h(nx)=nh(x) for all real numbers x and positive integers n satisfying nx∈(0,2/3). Thus, for any positive integers p and q, we have
h(qp​)=3ph(3q1​)=q3p​h(1/3)
which proves the claim when x is positive and rational.
Next, suppose for sake of contradiction that for some x∈(0,2/3), we have ∣h(x)−3h(1/3)x∣=δ>0. Consider any positive rational r<x. Then, we have by Claim 2 that
h(x−r)=h(x)−h(r)=h(x)−3h(1/3)r=h(x)−3h(1/3)x+3h(1/3)(x−r).
Thus, by taking r sufficiently close to x, we can ensure that
x−r<3⋅⌈1/δ⌉1​ and ∣h(x−r)∣>2δ​.
However, this implies (again by repeated applications of Claim 2)
∣h(2⋅⌈1/δ⌉⋅(x−r))∣=2⋅⌈1/δ⌉⋅∣h(x−r)∣>1,
which is a contradiction, since h must take values in (−1/3,2/3).
Thus, we have proved the claim for all positive x in the domain of h. Applying Claim 1, the result extends also to negative x, completing the proof.
By Claim 3, we conclude that h must take the form h(x)=cx, where c is a constant. Moreover, since h maps (−1/3,2/3) to itself, we must have c∈[−1/2,1]. In terms of f, this means we must have
f(x)=g(1/(x+1))=31​+c⋅(x+11​−31​)
for some constant −1/2≤c≤1. And we can readily check that all functions of this form do indeed work, by plugging this expression into the original equation, and choosing u,v,w such that x=vu​,y=wv​,z=uw​ as at the beginning of this solution (which can be done whenever xyz=1 ).
The problems on this page are the property of the MAA's American Mathematics Competitions