Prove that $4x^2-8xy+5y^2\geq0$ - is this a valid proof? I need to prove that $4x^2-8xy+5y^2\geq0$ holds for every real numbers $x, y$. 
First I start with another inequality, i.e. $4x^2-8xy+4y^2\geq0$, which clearly holds as it can be factorized into $(2x-2y)^2\geq0$. Now we can add $y^2$ (which is always nonnegative) to the left side, thus obtaining $4x^2-8xy+5y^2\geq0$, which is always true - we've just added a nonnegative expression to another, so the sum is still greater or equal to zero.
Is my proof correct? I had it on my final math exam and would like to be sure.

You gave a so called sum of squares certificate for the non-negativity. If you can write a polynomial as a sum of squares of polynomials, then it is non-negative, though the converse need not hold.  You may look up the history of "Hilbert's 17th problem" for more details.
