AM-GM might help (or may not) here

$\frac { { x }^{ 2 }+2xy+{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } \\ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { x }^{ 2 }+{ y }^{ 2 } } +\frac { 2xy }{ { x }^{ 2 }+{ y }^{ 2 } } \\ 1+\frac { 2xy }{ { x }^{ 2 }+{ y }^{ 2 } }$ From here, $\frac { 2xy }{ { x }^{ 2 }+{ y }^{ 2 } }$ , substitute x and y to sinA and cos A respectively. $\frac { 2sinAcosA }{ sin^{ 2 }A+{ cos }^{ 2 }A }$ Take note that $sin^{ 2 }A+{ cos }^{ 2 }A=1$ , and $2sinAcosA={ sin }2A$ , $\frac { { sin }2A }{ 1 } ,\quad -1\le { sin }2A\le 1$ The final answer is $1+1=\boxed{2}$

When solving problems like this one, I use AM-GM . So by simplifying, we get $\frac{(x+y)^{2}}{x^{2}+y^{2}}$ $=$ $\frac{x^{2}+y^{2}}{x^{2}+y^{2}}$ $+$ $\frac{2xy}{x^{2}+y^{2}}$ $=$ $1+\frac{2xy}{x^{2}+y^{2}}$ $(1)$ At this point, we use AM-GM to get an expression that we can substitute into $x^{2}+y^{2}$ . So, $\frac{x^{2}+y^{2}}{2}$ $\geq$ $\sqrt{x^{2}y^{2}}$ multiplying $2$ by both sides, we get $x^{2}+y^{2}$ $\geq$ $2xy$ $(2)$ . Lastly, by substituting $(2)$ into $(1)$ , we get $1+\frac{2xy}{2xy}$ , which gives $1+1$ $=$ $\boxed{2}$ as the answer. (Here, we are sure that we can use AM-GM because $x^{2}$ and $y^{2}$ are always positive. We can also say that both $x$ and $y$ can't simultaneously be equal to $0$ because if that happens, we will have an indeterminate form $\frac{0}{0}$ and we wouldn't be able to use AM-GM.)