Constrained optimisation: Minimize $(x+\frac{1}{x})^2 +(y+\frac{1}{y})^2$ subject to the constrain $x+y=1$ 
Minimize $(x+\frac{1}{x})^2 +(y+\frac{1}{y})^2$ subject to the
  constrain that $x+y=1$, where $x$ and $y$ are positive.

I used Lagrange-optimization and proved that 
$x-\frac{1}{x^3} = y - \frac{1}{y^3}.$
But I don't seem to get the answer. 
 A: Hint: Prove that $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq \frac{25}{2}$$
Your inequality is equivalent to
$$(x^2+y^2)(1+\frac{1}{(xy)^2})\geq \frac{17}{2}$$
Since we have $$x^2+y^2=1-2xy$$
we get
$$1-2xy+\frac{1}{(xy)^2}-\frac{2}{xy}\geq \frac{17}{2}$$
Substituting $$xy=t$$ the inequality above is equivalent
$$\frac{(1-4t)(2+4t+t^2)}{2t^2}\geq 0$$
This is true,since we get by AM-GM $\frac{1}{2}\geq \sqrt{xy}$
A: I didn't see a Lagrange multiplier approach in the linked answer so I think I should add this to the OP's answer:
The function $f(x) = x - {1 \over x^3}$ is strictly increasing as can be seen by taking its derivative and observing the result is positive. Strictly increasing functions are one-to-one, so if $f(x) = f(y)$ one has $x = y$. Since $x + y = 1$ here, this leads to $x = y = {1 \over 2}$, implying that $(x + {1 \over x})^2 + (y + {1 \over y})^2 \geq {25 \over 2}$. 
Note that to apply Lagrange multipliers here you have to work on domains $x, y > \epsilon, x + y \leq 1$ for small $\epsilon > 0$ and then let $\epsilon \rightarrow 0$.
A: Use Lagrange multipliers:
Let f=$(x+{1\over x})^2$+$(y+{1\over y})^2$ and let g=x+y
Then:
$\nabla$f=$(2x-{2\over x^3}$, $2y-{2\over y^3}$) and $\nabla$g=(1,1)
Solve the system equations:
$\nabla$f=$\lambda$$\nabla$g
g=1
Doing so implies that:
$(x-{1\over x^3}$ = $y-{1\over y^3}$) (1)
x+y=1 (2)
Substitute y=1-x into equation (1) to get:
$x-{1\over x^3}$ = $(1-x)-{1\over (1-x)^3}$
Solving the above equation yields 3 solutions, but the only one that satisfies all of the given constraints (x,y>$0$, x+y=1) is x=${1\over 2}$. Plugging this solution into (2) yields that y=${1\over 2}$.
When using Lagrange multipliers, we have to also check the points where either $\nabla$f is undefined or points where $\nabla$g=$0$ or is undefined. Both of these points still must satisfy the constraint g=1. Since $\nabla$g=(1,1), it is never $0$ or undefined. $\nabla$f is undefined at $(0,1)$ or $(1,0)$, but since f->$\infty$ as either x->$0$ or y->$0$, we ignore these solutions. 
Hence, the only constrained extrema we have is (${1\over 2}$,${1\over 2}$), and f (${1\over 2}$,${1\over 2}$)= ${25\over 2}$
All that is left to do is to prove that this constrained extremum is a minimum value of f.
In terms of an inequality, we must prove that:
$(x+{1\over x})^2$+$(y+{1\over y})^2$ <_${25\over 2}$ for all (x,y) s.t. x+y=1
I will refer you to the other answers posted above to prove this.
