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?
Best Answer
It is almost correct. The only flaw lies in the fact that you should assume only that $x\neq0$ or $y\neq0$. Therefore, you cannot say that both numbers $x^2$ and $y^2$ are greater than $0$. But you can say that their sum is greater than $0$, and that is quite enough for the proof.