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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 $0$

user21820
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Vons
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    How about $x=y=0$? – John B Jan 20 '17 at 22:49
  • 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. – dxiv Jan 20 '17 at 22:52

4 Answers4

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

Peter
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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? – Decaf-Math Jan 20 '17 at 22:56
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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 ? – Peter Jan 20 '17 at 23:02
  • Moreover, this representation does not need fractions (I admit, that it does not always work, but here, it does) – Peter Jan 20 '17 at 23:06
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Hint: Complete the square for the expression $10x^2-4xy$

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

DonAntonio
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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.

Nitin Uniyal
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  • I got the expression from the matrix. How do you know the matrix is positive definite? – Vons Jan 21 '17 at 02:29
  • 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. – Nitin Uniyal Jan 21 '17 at 03:42