Let's Prove A Nice Inequality

[@hexstealth7035]

Hello, can’t we use (a+b+c)^2 = 0 to prove 2(ac + ab + bc) = -(a^2 + b^2 + c ^2), so ac+bc+ab=-1/2(a^2+b^2+c^2) and since a,b and c are real numbers the RHS is negative or zero so the left is as well?

[@hitanshtiwari2784]

Valid

[@devotion7890]

Yes, the first method is already done at 0:55 because a sum of squares is greater or equal Zero, so to get Zero on the right side of the equation, the expression 2(ac + ab + bc) must be less or equal Zero and therefore also ac + ab + bc must be less or equal Zero

[@hexstealth7035]

@@devotion7890 yeah

[@mcwulf25]

Yes, and I thought that's what he was about to do.

[@ManjulaMathew-wb3zn]

Absolutely. That’s what I noticed at the first glance.

[@iMathematician]

( x+ y + z )^2 > 3 ( ab + bc + ca ) 0 > 3 ( ab…

[@robertveith6383]

Your post is wrong. For starters, there are no x, y, z variables.