Solution to puzzle 44: Sum of two powers 2

If x and y are positive real numbers, show that x^y + y^x > 1.

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Suppose x 1.
Then x^y 1 and y^x > 0.
Therefore x^y + y^x > 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) < y^x.

By symmetry, x/(x + y) < x^y.
Adding, we get 1 < x^y + y^x.

Therefore x^y + y^x > 1, for all positive real x and y.

Source: Problem 4 (page since taken down), from Shepherd College Math Club

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