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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.

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  • $\begingroup$ 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/… $\endgroup$ – Ethan Bolker Apr 1 at 17:26
  • $\begingroup$ Sorry if a,b,c,d are in the reals, would this hold? I believe that may be my problem $\endgroup$ – Rick Apr 1 at 17:28
  • $\begingroup$ 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. $\endgroup$ – Rick Apr 1 at 17:30
  • $\begingroup$ 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. $\endgroup$ – Ethan Bolker Apr 1 at 17:44
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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.

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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.

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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.

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