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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?

José Carlos Santos
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MrGreekMaths
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    It makes lot of sense: it is even correct. Nice! – DonAntonio Jul 17 '19 at 17:31
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    Cute. Most people would complete the square of $4x^2+6xy$ to obtain $4x^2+6xy+4y^2=(2x+3y/2)^2+7y^2/4,$ or would use the Quadratic Formula to show that $4x^2+6xy+4y^2=0$ implies both $x= (-6y\pm \sqrt {-28y^2})/8$ and $y=(-6x\pm \sqrt {-28x^2})/8.$ – DanielWainfleet Jul 18 '19 at 05:16

2 Answers2

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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.

José Carlos Santos
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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.

marty cohen
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