$xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
 A: Note that $y=1/x$ and $$\left(x+\frac{1}{x}\right)^2=\left( x-\frac{1}{x}\right)^2+4\cdot x\cdot \frac{1}{x}\geq4$$ Hence $x+y\geq 2$. Also for $x=y=1$, you have $x+y=2$, hence $2$ indeed the minimum value.
A: Arithmetic Mean $\ge$ Geometric Mean
$\frac{x+y}{2} \ge \sqrt{xy} $
$ \implies x + y \ge 2 $
$ \implies $ Minimum Value = $2$
A: The above answer does it totally.
However you can visualize this problem graphically. Imagine the three axes $x,y,z$ and the first quadrant part of the curve $xy=1$ drawn on the $xy$ plane. The function $f(x,y)=x+y$ is plotted on the $z$-axis. The graph of this function looks like a plane which is slanted 45% to the $xy$-plane and cuts it on the line $135^\circ$ to the $x$-axis. The minimum of this function with the constraint occurs where the curve $xy=1$ comes closest to the origin. You can see this by imagining the $z=x+y$ plane being parallely translated up and down.
Then it is clear that since a line of slope $135^\circ$ is tangent to the curve $xy=1$, this point of incidence will give you the minimum of $f$.
This point of incidence is $x=1,y=1$. and hence the minimum is $2$. And this picture also makes it clear that the minimum is attained.
A: $$x+y=\frac{x^2+1}{x}=\frac{x^2-2x+1}{x}+2 \geq0+ 2$$
A: $x+y=x+y-2\sqrt{xy}+2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2+2\ge2$
The value will be attained when $(\sqrt{x}-\sqrt{y})^2=0\Rightarrow \sqrt{x}=\sqrt{y}\Rightarrow x=y=1$
Please note that this is the basis for the inequality of A.M.$\ge$ G.M.
If $x,y\ge 0$ then we know that $(\sqrt{x}-\sqrt{y})^2\ge0$
$\Rightarrow x+y-2\sqrt{xy}\ge 0$
$\Rightarrow x+y\ge 2\sqrt{xy}$
A: This question has had a lot of answers, but none use what I would consider to be the easiest proof, so here's a general way to solve the problem:
$xy = 1$ can be re-written as $y = {1\over x}$, and putting that back into $x+y$ we get:
$$x+y=x+{1\over x}=x+x^{-1}$$
which differentiates to:
$${d\over dx} (x+x^{-1})=1-x^{-2}$$
Now at a turning point this will be 0. That's at $x=\pm 1$. So the two turning points are at $x=-1,y=-1$ and $x=1,y=1$.
You already restricted the solution to positive numbers (them being real makes no difference), but for completeness we still need to look at the edge cases of $x=\inf$ and $x=0$ (which both produce infinite results).
Instead of taking edge cases, you could also show that this is a minimum (rather than maximum or point of inflection) by taking the second differential and checking the sign:
$${d^2\over dx^2} (x+x^{-1})=2x^{-3}$$
At $x=1$, this is $2$. Since this is greater than 0, we know it is a minimum. Note that in this case it is the only turning point within the range, so must be a global minimum, but in general you would have to check every turning point.
So your minimum is $2$, at $x=1, y=1$.
Some references if you're new to derivatives:


*

*http://www.mathsrevision.net/alevel/pages.php?page=45

*http://www.mathsrevision.net/alevel/pages.php?page=43
A: With $x\cdot y=1$ you know that $x=\frac{1}{y}$ so 
$$x+y=x+\frac{1}{x}$$ 
Wlog you can say now $x\geq 1 $, try to find the minimum now.
Another way is for x>0 we know that 
$$\frac{(x-1)^2 }{x}\geq 0$$ 
on the other hand 
$$\frac{(x-1)^2}{x} = \frac{x^2}{x} - \frac{2x}{x} +\frac{1}{x} 
=x-2 +\frac{1}{x} \geq 0$$
This is equivalent to $$x+\frac{1}{x}\geq 2$$ 
and as $1+1=2$ we know that 2 is the minimum.
I guess I get some downvotes for the following, but maybe it gives you some hope. 
If you don't like adding a smart zero (which is a bit unintuitive), you can still use advanced calculus methods.  You try to minimize the function $f:\mathbb{R}^2\rightarrow \mathbb{R}; \ (x,y)\mapsto x+y$ with the side condition $x\cdot y = 1$. 
Now you use a lagrangian multiplier and minimizing the function $$g(x,y)=x+y+\lambda (xy-1)$$ 
Now you take partial derivatives, which must be zero for a minimum
\begin{align*}
\frac{\partial g}{\partial x} &= 1+ \lambda y\\
\frac{\partial g}{\partial y} &= 1+ \lambda x\\
\frac{\partial g}{\partial \lambda} &= xy-1
\end{align*}
Now you only need to solve the system 
\begin{align*}
0&=1+\lambda y\\
0&= 1+\lambda x \\
0&=xy - 1 
\end{align*}
As Winston mentioned you could just take the derivative of $\displaystyle x+\frac{1}{x}$  which is 
$$1-\frac{1}{x^2}$$
So you need to solve the equation 
$$1-\frac{1}{x^2}=0 \iff 1=\frac{1}{x^2} \iff x^2=1$$
A: Here is another approach. We have $ x, y > 0 $ and 
$$ xy=1 \implies y=\frac{1}{x}. $$
Substituting $ y=\frac{1}{x} $ in the equation $f(x,y)=x+y$ yields 
$$ F(x)= x+\frac{1}{x}.$$
Now, we have a function in one variable which can be minimized using techniques from univariate calculus as
$$  F(x)= x+\frac{1}{x} \implies F'(x)=1-\frac{1}{x^2}=0\implies x = 1>0. $$
$$ \implies F''(1)=2>0 $$
which means $x=1$ is a minima for the function $F(x)$ and the minima is $F(1)=2.$
So the minimum of the function $f(x,y)$ attains at the point $ (x,y)=(1,1) $.
A: I agree with magguu about visualizing the problem graphically, but I'd do the visualization in two dimensions rather than three.  In the $x$-$y$ plane, imagine the graph of $xy=1$ (a hyperbola) and imagine the level lines of $x+y$ (lines sloping downward, making $45^\circ$ angles with the axes).  In more detail, start with the level line through the origin ($xy=0$), which misses the hyperbola entirely, and imagine gradually moving this line to larger values of $x+y$, until it first touches the hyperbola.  By symmetry between $x$ and $y$ (and convexity of the hyperbola), this happens at the point on the hyperbola where $x=y$, i.e., at the point $(1,1)$.
A: $$(\sqrt x-\sqrt y)^2\geq 0$$
$$x-2\sqrt{xy}+y\geq 0$$
$$x+y\geq 2\sqrt{xy}$$
$$x+y\geq 2\cdot 1$$
$$x+y\geq 2$$
A: If for  $c>0, xy=c^2\implies y=\frac{c^2}x\implies x+y=x+\frac{c^2}x=z$(say)
So, $x^2-zx+c^2=0$ 
As this is a quadratic equation in $x$ and $x$ is real,
the discriminant  $(-z)^2-4\cdot1\cdot c^2\ge 0\implies z^2\ge4c^2\implies $ either $z\ge2c$ or $z\le -2c$
As $x,y>0,x+y>0$ 
So,  $x+y=x+\frac1x=z\ge2c$
A: Consider a rectangle with dimension (x, y) of unit area, the half perimeter is defined as $a+b$
a square is a one and only one special form of rectangle with equal dimension should either be a maxima or minima
Now lets calculate the half perimeter of a square
$x.y = 1$
as $x = y$
we have
$x.x = 1$
so $x = 1$
which gives us 
$x + y = 2$
Consider any one rectangular figure
$x = 2$ 
and 
$y = \frac{1}{2}$
this gives the perimeter as
$x + y = 2 + 1/2 = 2.5$
$2.5(rectangle) > 2 (square)$
which makes us to believe
a square will have the minimum perimeter
thus we can conclude
the minimum value of $x + y$ is $2$
A: as $xy=1$;
so, $y=1/x$;
as we know that 
$(x+y)^2 = x^2+y^2+2xy$;
$(x+y)^2 = x^2 = 1/x^2 +2$;
so, $x+y = \sqrt{x^2 + \frac{1}{x^2} + 2}$
to be minimum, compare RHS by 0;
you will get $x=-1$; but as $x$ is a +ve real number, 
so, $x$ is minimum at 1;
so $x+y =2$;
