Prove that $x^2 + 5xy+7y^2 > 0$ for all $x,y \in\mathbb{R}$

Question

This is probably really easy for all of you, but my question is how do I prove that $x^2 + 5xy+7y^2 > 0$ for all $x,y\in\mathbb{R}$

Thanks for the help!

Answer 1

$$x^2+5xy+7y^2=\left(x+\frac{5y}2\right)^2 + \frac{3y^2}4\ge 0$$ (not $>0$ as $x=y=0$ leads to 0).

Answer 2

You can’t prove that it’s always greater than $0$, because when $x=y=0$ it is $0$, but you can prove that it’s always greater than or equal to $0$ by completing the square:

$$x^2+5xy+7y^2=\left(x+\frac{5y}2\right)^2+\frac34y^2\;.$$

Since square are always at least $0$, so is their sum.

Answer 3

Another possibility is to note that the quadratic form is represented by the matrix

$$ \begin{pmatrix} 1 & 2.5 \\ 2.5 & 7\end{pmatrix} $$ The claim follows from the fact that its eigenvalues are both positive, so that the matrix is positive definite.

Answer 4

Using A.M.-G.M. inequality, $$x^2 + 7y^2 \geq 2 \sqrt{x^2 7 y^2} = 2 \sqrt 7 |x|\cdot |y| \geq 5 |x| \cdot |y| \geq -5 xy \implies x^2 + 5 xy + 7y^2 \geq 0$$

Answer 5

Another approach, playing a little with algebra-geometry: as $\,x\neq 0\,\,or\,\,y\neq 0\,$ , since otherwise, as others pointed out, you have equality to zero, assume $\,y\neq 0\,$ and look at the expression as quadratic (= a parabola) in $\,x\,$ :

$$x^2+(5y)x+7y^2\Longrightarrow \Delta:=b^2-4ac=25y^2-28y^2=-3y^2< 0$$

Thus, the quadratic keeps one single sign all the time (otherwise the parabola would intersect the x-axis), and since at $\,x=0\,$ we get the value $\,7y^2>0\,$ we have then that the expression is always positive (i.e., the parabola is always over the x-axis)