Proof of inequality problem from Spivak, ch 1, 16 b) If $$4x^2+8xy+4y^2 \ge 0$$ it follows that $$4x^2+6xy+4y^2 \gt 0 $$ unless $x=0$ and $y=0$. 
How can I prove that?
 A: First, the premise is uninformative because you always have
$$
0\leq (2x+2y)^2=4x^2+8xy+4y^2.
$$
Second, it is also unnecessary because you also always have
$$
4x^2+6xy+4y^2=(2x+3y/2)^2+7y^2/4\geq 0
$$
with equality iff $x=y=0$.
A: Hint:$$4x^2+6xy+4y^2=x^2+3(x+y)^2+y^2\ .$$
A: hint:$$4x^2+6xy+4y^2 = 4\left(x^2+\dfrac{3}{2}xy+y^2\right)=4\left(\left(x+\dfrac{3}{4}y\right)^2+\dfrac{7}{16}y^2\right)$$
A: The way that I think of it is as follows. First, we need to recognise that
$$ 4x^2+6xy+4y^2=3(x+y)^2 + x^2 + y^2$$
Then, based on the stated assumption we know that:
$$ 4x^2+8xy+y^2=4(x+y)^2 \ge 0 $$
$$ (x+y)^2\ge0$$
$(x+y)^2 = 0$ when $x=-y$ or when $x=y=0$
For all other values $(x+y)^2\gt0$
Similarly, 
$$ x^2 + y^2 \ge 0$$
$x^2 + y^2 = 0$ where $x=y=0$
For all other values, $x^2+y^2\gt0$
Combining these two steps we find that where $x$ and $y\not=0$:
$$3(x+y)^2\ge0\; and\; x^2 + y^2 \gt0$$
$$\therefore\;3(x+y)^2+x^2+y^2\gt0$$
$$4x^2+6xy+4y^2\gt0$$
P.S. I'm just working through this chapter myself, so my understanding is pretty basic. I'm sure others have expressed this more elegantly, but hopefully my fumbling thought process will be helpful for other beginners.
