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

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    How about $x=y=0$? – 2017-01-20
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    Hint: assume WLOG that $y \ne 0$. Then write it as $y^2\big(10(x/y)^2-4(x/y)+30\big)\,$ where the latter factor is a quadratic in $x/y\,$ whose sign is easy to determine. – 2017-01-20

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

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    This answer is perfectly acceptable, but it leaves the reader with the question regarding "why?". Moreso, what's the motivation that led you to see this versus some other "standard" argument like completing the square? – 2017-01-20
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    It is not difficult to find out that we need the terms $x$ and $2y$ to produce the crucial $4xy$-term, so what is the mysterical part of my calculation ? – 2017-01-20
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    Moreover, this representation does not need fractions (I admit, that it does not always work, but here, it does) – 2017-01-20
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Hint: Complete the square for the expression $10x^2-4xy$

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

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$q(x,y)=10x^2+30y^2-4xy$ 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.

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    I got the expression from the matrix. How do you know the matrix is positive definite? – 2017-01-21
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    One way to see it that eigenvalues are positive. Another way to see that minors $|a_{11}|$ and $|A|$ are positive. Here these minors are $10$ and $296$ respectively. – 2017-01-21