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,

<span> x^y + y^x > 1

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

I hope it has come to your help.

</span>

6 0

4 years ago

If the Revenue for the week is $2000, and labor cost consists of two workers earning $8 per hour who work 40 hours each, what is

Ira Lisetskai [31]

Let's find out the cost of labor shall we...

$8 x 40 x 2 = $640

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Now let's answer your question...

($640)/($2000) = 0.32

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This means that the cost of labor would be equal to 32% of the revenue as 0.32 = 32/100.

6 0

4 years ago

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