1. ## Algebra

Given two real numbers $x$ and $y$ such that
$(x+y)^2\leq2xy$
find $x+y$.

If $x+y=6$ and $xy=3$, find $\displaystyle\frac{1}{x}+\frac{1}{y}$.

Given $|x-y|=2$ and $|y-z|=7$, find the minimum possible value of $|x-z|$.

2. 1. $(x+y)^{2} \leq 2xy$
$x^{2} + 2xy + y^{2} \leq 2xy$
$x^{2} + y^{2} \leq 0$
$x^{2} \leq -y^{2}$

What values of x and y make this inequality even possible?

2. $\frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x + y}{xy}$

3. $|x - y| = 2 \quad \Rightarrow \quad x - y = \pm 2 \quad \Rightarrow \quad x = y \pm 2$
$|y - z| = 7 \quad \Rightarrow \quad y - z = \pm 7 \quad \Rightarrow \quad z = y \pm 7$

Find x - z and see what the lowest value of |x - z| would be.

3. Originally Posted by o_O
1. $(x+y)^{2} \leq 2xy$
$x^{2} + 2xy + y^{2} \leq 2xy$
$x^{2} + y^{2} \leq 0$
$x^{2} \leq -y^{2}$

What values of x and y make this inequality even possible?
$x^2+y^2 \leq 0$. Note $x^2+y^2\geq 0$ (since it is a sum of two squares). Thus, $x^2+y^2 = 0 \implies x=0,y=0$.