1. ## more inequalities

show that x^2 - 4xy + 5y^2 >= 0 for all real values of x and y.
to do this must i simply prove that the quardatic has no real roots? does the absence of real roots always prove that all values of the given variable fulfill the conditions?

2. $\displaystyle x^2-4xy+5y^2$
$\displaystyle =x^2-4xy+4y^2+y^2$
$\displaystyle =(x-2y)^2+y^2\geq0$
(being the sum of 2 squares)

3. Hello, furor celtica!

Show that: .$\displaystyle x^2 - 4xy + 5y^2 \:\geq\:0$ for all real values of $\displaystyle x$ and $\displaystyle y.$

We have: .$\displaystyle x^2 - 4xy + 4y^2 + y^2 \:=\:(x -2y)^2 + y^2$

The square of any real number is nonnegative.
And the sum of two nonnegative numbers is nonnegative.

Therefore: .$\displaystyle (x-2y)^2 + y^2 \:\geq \:0$

4. the absence of real roots indicates that x^2 - 4xy + 5y^2 > 0 not x^2 - 4xy + 5y^2 >= 0 because if it does = 0 then there are roots. Btw I might be wrong

5. can somebody verify this?
that was actually my main question, i know that it can be proved using perfect squares but i was wondering abouit the root part.