Thread: 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?

(being the sum of 2 squares)

Hello, furor celtica!

Show that: . for all real values of and

We have: .

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

Therefore: .

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

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.