If x and y are positive real numbers, show that xy + yx > 1.
Suppose x 1.
Then xy 1 and yx > 0.
Therefore xy + yx > 1.
Now suppose 0 < x < 1. Then 1/x > 1.
By Bernoulli's inequality (Mitrinovic's generalization of), (1 + x/y)1/x > 1 + 1/y.
Hence (1 + x/y)1/x > 1/y.
Then 1 + x/y > (1/y)x, since x/y > 0 and y > 0.
That is, (x + y)/y > (1/y)x.
Taking the reciprocal of both sides reverses the inequality, and so y/(x + y) < yx.
By symmetry, x/(x + y) < xy.
Adding, we get 1 < xy + yx.
Therefore xy + yx > 1, for all positive real x and y.
Source: Problem 4 (page since taken down), from Shepherd College Math Club