XLIX OM - I - Problem 4

Given a positive number $ a $. Determine all real numbers $ c $ that have the following property: for every pair of positive numbers $ x $, $ y $, the inequality $ (c-1)x^{a+1}  \leq (cy - x)y^a $ is satisfied.
Assume that a real number $ c $ has the considered property:

Substituting $ y=tx $ and dividing both sides by $ x^{a+1} $, we obtain the relation

Notice that the converse is also true: condition (2) implies (1). Suppose that $ c $ is a real number satisfying condition (2). Let $ x $, $ y $ be any positive numbers. In inequality (2), substitute $ t = y/x $, multiply both sides by $ x^{a+1} $, and we get inequality (1).
Therefore, conditions (1) and (2) are equivalent. Fix a real number $ c $ and consider the function $ f:(0,\ \infty)\to \mathbb{R} $ given by

of course $ f(1) = c - 1 $ and

If condition (2) (equivalent to condition (1)) is satisfied for the considered number $ c $, then the function $ f $ has a minimum at the point $ t = 1 $, so $ f'(1) = 0 $, which means $ c(a+1)-a= 0 $; hence $ c=\frac{a}{a+1} $.
Conversely, if in the definition of the function $ f $ we take $ c= \frac{a}{a+1} $, then

therefore, the function $ f $ is decreasing in the interval $ (0;\ 1 \rangle $ and increasing in the interval $ \langle 1;\ \infty) $. At the point $ t = 1 $, it therefore attains its minimum value. In other words, $ f(t)\geq f(1) $ for $ t>0 $, which means that condition (2) (equivalent to condition (1)) is satisfied.
Conclusion: for a given positive number $ a $, the only number $ c $ having the postulated property is $ c=\frac{a}{a+1} $.