How to prove $\frac xy + \frac yx \ge 2$ I am practicing some homework and I'm stumped.
The question asks you to prove that
$x \in Z^+, y \in Z^+$
$\frac xy + \frac yx \ge 2$
So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy
This is my proof so far for opposite parity
$x,y \in Z^+ $ $|$ $x \gt 0,$ $y \gt 0$. Let x be even $(x=2a, $ $ a \in Z^+)$ and y be odd $(y=2b+1, $ $b \in Z^+)$. Then,
$\frac xy + \frac yx \ge 2$
$\frac {2a}{2b+1} + \frac {2b+1}{2a} \ge 2$
$\frac {2b^2 + 4a^2 + 4a + 1}{2b(2a+1)} \ge 2$
$4b^2 + 4a^2 +4a + 1 \ge 4b(2a+1)$
$4b^2 + 4a^2 + 4a +1 \ge 8ab + 4b$
$4b^2 - 4b + (2a + 1)^2 \ge 8ab$
$(2b-1)^2 + 1 + (2a+1)^2 \ge 8ab$
I feel like this is the not the correct way to go about proving it, but I can't think of a better way to do it. Does anyone have any suggestions? Just hints please, not a solution. 
 A: Clearly $(x-y)^2 \geq 0$
So $x^2-2xy+y^2 \geq 0 \Rightarrow x^2+y^2 \geq 2xy $
Since $x $ and $y$ are positive integers , $x \neq 0 $ and $y \neq 0$.
Thus we can divide by $xy$.
So we have $\frac{x^2+y^2}{xy} \geq \frac {2xy}{xy} \Rightarrow \frac {x}{y}+\frac{y}{x} \geq 2$.
A: Hint: Consider multiplying both sides of the inequality by $xy$, which is positive. Then, we have:
$$\frac xy + \frac yx \ge 2\\
x^2+y^2\ge 2xy
$$
Can you go on?
A: $$(x-y)^2\geq0$$

 $$\begin{array}{l}x^2-2xy+y^2\geq0\\x^2+y^2\geq2xy\\\frac{x^2+y^2}{xy}\geq2\\\frac{x^2}{xy}+\frac{y^2}{xy}\geq2\\\frac{x}{y}+\frac{y}{x}\geq2\\QED\end{array}$$

This will work for $x,y\in\mathbb{R}\backslash\{0\}$
A: For every $a$ you have $(a-1)^2\geq 0$, so $a^2+1\geq 2a$. 
If $a>0$ you have $a+\frac 1a\geq 2.$ Take $a=\frac xy$.
A: If $x=y,$ then it is true.
Suppose $x>y$ then there is exists positive integer $n$ such that $x=y+n.$
This implies $\frac{x}{y}=1+\frac{n}{y}$ and $\frac{y}{x}=1-\frac{n}{x}$.
Adding these equations, we obtain
\begin{align}
\frac{x}{y}+ \frac{y}{x}=2+n\left(\frac{1}{y}-\frac{1}{x}\right)=2+n\frac{x-y}{xy}> 2\quad as\quad x>y.
\end{align}
A: Consider $f(t) = t+{1 \over t}$ on $t >0$. $f$ is convex and $f'(1) = 0$, hence $1$ is a global maximiser. Hence $f(t) \ge f(1) = 2$ for all $t > 0$.
Hence ${x \over y} + { y \over x} = f({x \over y}) \ge 2$ for all $x,y>0$.
A: You answer is correct but you went the wrong way near the end. (Plus you have  minor error in it (you swapped a and b somewhere which doesn't really change the problem).
You had: 
$\frac{x}{y}+\frac{y}{x}\geq2$
$\frac{2a}{2b+1}+\frac{2b+1}{2a}\geq2$
$\frac{4a^2+4b^2+4\color{red}b+1}{2\color{red}a(2\color{red}b+1)}$
$4a^2+4b^2+4b+1\geq4a(2b+1)$
$4a^2+4b^2+-8ab-4a+4b+1\geq0$
$(2a+2b-1)^2\geq0$
A: Here is "another" simple method(this is essentially same as completing the square.) of proving $\frac{x}{y}+\frac{y}{x}\geq 2$ by the help of AM-GM Inequality :
Consider the set $\{\frac{x}{y},\frac{y}{x}\}.$
Applying AM-GM Inequality on these two values , we have :
$$\frac{\frac{x}{y}+\frac{y}{x}}{2} \geq \sqrt {\frac {x}{y} \times \frac{y}{x}}$$
$$\implies \frac{\frac{x}{y}+\frac{y}{x}}{2} \geq \sqrt {1}$$
$$\implies \frac{x}{y}+\frac{y}{x} \geq 1 \times 2$$
$$\implies \frac{x}{y}+\frac{y}{x} \geq 2$$
Hople this helps. :)
A: Without loss of generality, we assume $t:=\frac{x}{y}\geq 1,$ we see that the $t+\frac{1}{t}$ has its derivative non-negative, so $t+\frac{1}{t}\geq 2.$
A: By rearrangement inequality
$$\frac xy + \frac yx = x\cdot \frac 1 y + y \cdot \frac 1x\ge x\cdot \frac 1 x + y \cdot \frac 1y=1+1=2$$
