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
----------------------------------------------------
I have to give credit to my colleague, Mikhael Glen Lataza for the wonderful solution.
I hope it has come to your help.
</span>
Answer:
1728
Step-by-step explanation:
<h2>
<em>OladipoSeun</em><em>♡˖꒰ᵕ༚ᵕ⑅꒱</em></h2>
The expected value is
![E[X]=\displaystyle\sum_xx\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x=\dfrac{0+1+\cdots+9+10}{11}=\dfrac{55}{11}=\boxed{5}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Csum_xx%5C%2CP%28X%3Dx%29%3D%5Cfrac1%7B11%7D%5Csum_%7Bx%3D0%7D%5E%7B10%7Dx%3D%5Cdfrac%7B0%2B1%2B%5Ccdots%2B9%2B10%7D%7B11%7D%3D%5Cdfrac%7B55%7D%7B11%7D%3D%5Cboxed%7B5%7D)
The standard deviation is the square root of the variance, which is
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
where
![E[X^2]=\displaystyle\sum_xx^2\,P(X=x)=\frac1{11}\sum_{x=0}^{10}x^2=\dfrac{0^2+1^2+\cdots+9^2+10^2}{11}=\dfrac{385}{11}=35](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Csum_xx%5E2%5C%2CP%28X%3Dx%29%3D%5Cfrac1%7B11%7D%5Csum_%7Bx%3D0%7D%5E%7B10%7Dx%5E2%3D%5Cdfrac%7B0%5E2%2B1%5E2%2B%5Ccdots%2B9%5E2%2B10%5E2%7D%7B11%7D%3D%5Cdfrac%7B385%7D%7B11%7D%3D35)
so that
![V[X]=35-5^2=10](https://tex.z-dn.net/?f=V%5BX%5D%3D35-5%5E2%3D10)
making the standard deviation
![\sqrt{V[X]}=\sqrt{10}\approx\boxed{3.16}](https://tex.z-dn.net/?f=%5Csqrt%7BV%5BX%5D%7D%3D%5Csqrt%7B10%7D%5Capprox%5Cboxed%7B3.16%7D)
