<s>[INST] How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$ 
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$ 
Due to the fact that  $(x+y)^2$ is a square, it will be positive 
$x>0$ and $y>0$ so $xy>0$ 
So $\frac{(x+y)^2}{xy}>0$
So $\frac{x}{y}+\frac{y}{x}>2$ 
But the problem is that I have proven that $\frac{x}{y}+\frac{y}{x}>2$, but in the case of $x=y$, it is equal, so it has to be $\geq$...
Could someone help?
 [/INST] Let $a = x/y$, then this is equivalent to proving $a + 1/a \geq 2$. $a$ is positive so multiply both sides by $a$ we get
$$a^2 - 2a + 1 = (a - 1)^2 \geq 0.$$
</s>