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