<s>[INST] Proof Verification: $4x^2+6xy+4y^2>0$ unless both $x$ and $y$ are equal to $0$ This is a problem from Spivak's Calculus 4th ed., Chapter 1, Problem 16(b).

Using the fact that
  $x^2 + 2xy + y^2 = (x + y)^2 > 0$,
  show that $4x^2 + 6xy + 4y^2 >0$ unless $x$ and $y$ are both $0$.

This is my proof:
Firstly, as any non-zero number $a$ satisfies the inequality $a^2>0$ (I have proved that already), $(x+y)^2>0 \Rightarrow 3(x+y)^2$, $x^2>0$ and $y^2>0$
$\therefore 3(x+y)^2+x^2+y^2=4x^2+6xy+4y^2>0$
Does this make sense?
 [/INST] More generally,
since
$\begin{array}\\
ax^2+2bxy+ay^2
&=a(x^2+y^2)+b(x^2+2xy+y^2)-b(x^2+y^2)\\
&=(a-b)(x^2+y^2)+b(x+y)^2\\
\end{array}
$
if
$a > b$ then
$ax^2+2bxy+ay^2 \ge 0$
with equality if and only if
$x = y = 0$.
Your case is
$a=4, b=3$.
You can handle the more general case of
$ax^2+2bxy+cy^2$
by looking at how
the quadratic formula
is derived.
</s>