I am trying to prove $a^2-4ab+b^2=0$ has no solutions for all real numbers $a$ and $b$ and $b \neq 0$
We know that $a^2 \geq 0$ and $b^2 > 0$ since $b \neq 0$. So then $a^2 + b^2 >0$. Now I'm stuck as I'm not sure how to show that $a^2 + b^2 = 4ab$ has no solutions given the above conditions. A little assistance would be greatly appreciated.