Show that $10x^2+30y^2-4xy > 0$ when $x,y$ are not both zero

Question

Show that $10x^2+30y^2-4xy$ is greater than 0 for all values of $x,y$ When $x,y$ are not both $0$

Answer 1

Hint : $$10x^2+30y^2-4xy=(x-2y)^2+9x^2+26y^2$$

Answer 2

$$10x^2-4xy+30y^2=10\left(x-\frac y5\right)^2-\frac{2y^2}{5}+30y^2=10\left(x-\frac y5\right)^2+\frac{148}5y^2$$

Answer 3

Hint: Complete the square for the expression $10x^2-4xy$

Answer 4

$q(x,y)=10x^2+30y^2-4xy$ is equivalent to $\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}10&-2\\-2&30\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$. Now see the matrix is positive definite.