Maximize the expression $\frac{1}{x}+\frac{1}{y}$, given that $\frac{1}{x^n}+\frac{1}{y^n}<1$, for natural numbers $x,y,n$ $x$, $y$ are natural numbers such that $\dfrac{1}{x^n}+\dfrac{1}{y^n}<1$. Maximise $\dfrac{1}{x}+\dfrac{1}{y}$.
Here $n$ is a natural number also.
 A: You can try using Lagrangian multiplies to find stationary points (saddle points, maxima, minima). As Macavity suggested we gonna split the problem into 2 sub-problems.
Case 1: $n \geq 2$
The function is: $f(x,y) = \frac{1}{x^n} + \frac{1}{y^n} = x^{-n} + y^{-n}$. And we have two constraints those are: $x,y \geq 2$, because otherwise the function will violate the condition $\frac{1}{x^n} + \frac{1}{y^n} < 1$
$$f(x,y) =  x^{-n} + y^{-n}$$
$$g(x) = x \geq 2$$
$$h(y) = y \geq 2$$
$$F(x,y,\lambda,\lambda_1) = x^{-n} + y^{-n} + \lambda(x-2) + \lambda_1(y-2)$$
Now we take partial derivatives:
$$F_x = -nx^{-n-1} + \lambda = 0$$
$$F_y = -ny^{-n-1} + \lambda_1 = 0$$
Beacuse the two last terms has to be equal to zero we have 4 cases.
Cases $\lambda = \lambda_1 = 0$ and $\lambda_1 = 0,x = 2$ and $\lambda = 0, y = 2$ wouldn't work, beacuse that would imply that one of the variables is 0, which isn't possible. That leaves us with only one option: $x = y = 2$
Now checking into the initial function for $(x,y) = (2,2)$ and for a random n, we get that this point is maxima of the function.
So the maximum of the function occurs at $(2,2)$ and it's $f(2,2) = \frac{1}{2^{n-1}}$
Case 2: $n = 1$
The function now looks like this:
$f(x,y) = x^{-1} + y^{-1}$ and we have constraints the same two constraints and an additional one, which is: $\frac{x+y}{xy} < 1$ 
$$f(x,y) =  x^{-1} + y^{-1}$$
$$g(x) = x \geq 2$$
$$h(y) = y \geq 2$$
$$j(x,y) = x+y \leq xy - 1$$
$$F(x,y,\lambda,\lambda_1, \lambda_2) = x^{-1} + y^{-1} + \lambda(x-2) + \lambda_1(y-2) + \lambda_2(x+y+1-xy)$$
Now we take partial derivatives:
$$F_x = -x^{-2} + \lambda + \lambda_2 - \lambda_2y = 0$$
$$F_y = -y^{-2} + \lambda_1 + \lambda_2 - \lambda_2x = 0$$
Because the last 3 terms have to be equal to 0, we have 8 distinct cases:
Sub-case 1: $\lambda = \lambda_1 = \lambda_2 = 0$
This method doesn't give a solution, because it implies $x=y=0$, which is impossible.
Sub-case 2: $\lambda = \lambda_1 = x+y-xy+1 = 0$
From $x+y-xy+1 = 0$ we obtain:
$$y+1 = xy-x \implies x(y-1) = y+1 \implies x = \frac{y+1}{y-1}$$
Beacuse x is an integer, this implies $y=2$ and $x=3$.
Also we have:
$$x+1 = xy-y \implies y(x-1) = x+1 \implies y = \frac{x+1}{x-1}$$
Beacuse y is an integer, this implies $x=2$ and $y=3$.
So in this case we obtained two solutions: $$(x,y) = (2,3), (3,2)$$
Sub-case 3: $\lambda = y-2 = \lambda_2 = 0$
This implies $(0,2)$ as solution which is imposible.
Sub-case 4: $\lambda = y-2 = x+y-xy+1 = 0$
This implies a solution that we've already obtained, which is $(3,2)$.
Sub-case 5: $x-2 = \lambda_1 = \lambda_2 = 0$
This implies $(2,0)$ as solution which is imposible.
Sub-case 6: $x-2 = \lambda_1 = x+y-xy+1 = 0$
This implies a solution that we've already obtained, which is $(2,3)$.
Sub-case 7: $x-2 = y-2 = \lambda_2 = 0$
This implies a solution $(2,2)$, which violate the third constrain and is impossible.
Sub-case 8: $x-2 = y-2 = x+y-xy+1 = 0$
This is a contradiction, because it's impossible these 3 constraints to be equal to 0.
Because we exhausted all the sub-cases, we check the two solution we obtained and we get that they are global maxima of the function.
Conclusion
If $n \geq 2$ then the maximum occurs at point $(2,2)$, otherwise it occurs at points $(3,2)$ and $(2,3)$.
