# Simple math question that seems intuitive

Let $$a,b,c,d$$ be real numbers.

$$3(a^2+b^2)+2(c^2+d^2)+2(ab+cd)$$

How can one show that this is always greater than zero, unless $$a=b=c=d=0$$.

I have to show a positivity axiom for an inner product and this is what I'm left with. Hence why i expect it to be greater than zero.

• It's not true when $a=i$ and $b=c=d=0$. The problem may be that I can't read or you have misentered the expression. Please edit to use mathjax: math.meta.stackexchange.com/questions/5020/… – Ethan Bolker Apr 1 at 17:26
• Sorry if a,b,c,d are in the reals, would this hold? I believe that may be my problem – Rick Apr 1 at 17:28
• Im expressing complex numbers as a+bi, c+di ie.) the complex numbers are R^2 equipped. So the values aboves are obviously real numbers. I’m so happy you pointed this out thank you. – Rick Apr 1 at 17:30
• You're welcome. Thanking me in a comment is not enough. You should edit the question so that it's correct and use mathjax to format the mathematics. – Ethan Bolker Apr 1 at 17:44

## 3 Answers

This is equal to $$2a^2+2b^2+c^2+d^2 +(a+b)^2+(c+d)^2$$ which is obviously positive for $$a,b,c,d\in\mathbb{R}$$ but is not necessarily so otherwise.

This is a quadratic form... Just analise it using eigenvalues or the determinants of the principal minors of the associated matrix. If you conclude that the matrix is positive definite, your conclusion holds.

As previous answers already showed, This is also equivalent to (a+b)^2+(c+d)^2+a^2+c^2+b^2+d^2 That is always positive for any a, b, c, d € R, Only 0 if all of them are zero.