Proof that x^y + y^x > 1 for all x, y > 0

To prove this inequality we need to consider three cases. We need to see that the equation is symmetric and that switching the variables x and y does not change the equation.

Case 1: x >= 1, y >= 1

It is obvious that

 x^y >= 1, y^x >= 1
 x^y + y^x >= 2 > 1
 x^y + y^x > 1

Case 2: x >= 1, 0 < y < 1

 Considering the following sub-cases:

  - x = 1, x^y = 1
  - x > 1,

    Let x = 1 + n, where n > 0

    x^y = (1 + n)^y = f_n(y)

    By Taylor Expansion of f_e(y) around y = 0,

    x^y = f_n(0) + f_n'(0)/1!*y + f_n''(0)/2!*y^2 + ...
        = 1 + ln(1 + n)/1!*y + ln(1 + n)^2/2!*y^2 + ...

    Since ln(1 + n) > 0,

    x^y > 1

  Thus, we can say that x^y >= 1, and since y^x > 0.

  x^y + y^x > 1

  By symmetry, 0 < x < 1, y >= 1, also yields the same.

Case 3: 0 < x, y < 1

  We can prove this case by fixing one variable at a time and by invoking symmetry to prove the relation.

  Fixing the variable y, we can set the expression as a function,

  f(x) = x^y + y^x
  f'(x) = y*x^(y-1) + y^x*ln y 
  For all x > 0 and y > 0, it is obvious that
  f'(x) > 0.

  Hence, the function f(x) is increasing and hence the function f(x) would be at its minimum when x -> 0+ (this means close to zero but always greater than zero). 

  lim x->0+ f(x) = 0^y + y^0 = 0 + 1 = 1

  Thus, this tells us that 

  f(x) > 1.

  Fixing variable y, by symmetry also yields the same result: f(x) > 1.

  Hence, when x and y are varying, f(x) > 1 must also hold true.
    Thus, x^y + y^x > 1.

We have exhausted all the possible cases and shown that the relation holds true for all cases. Therefore, 

  x^y + y^x > 1



I have to give credit to my colleague, Mikhael Glen Lataza for the wonderful solution.

I hope it has come to your help.

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