Question

Prove that the inequality is true for all real numbers a,b,c:

Answer 1

\[ 4(a^2 + b^2 + c^2) - \big[ (a+b)^2 + (b+c)^2 + (a+c)^2\big] \ge 0\]

Answer 2

I got \( a^2 + b^2 + c^2 -ab - bc - ac \ge 0 \). But I don not know how to continue.

Answer 3

I thought that I can somehow start with the fact that \((a+b+c)^2 \ge 0\).

Answer 4

\[\Large a^2+b^2+c^2 \geq ab+bc+ac \] or \[\Large2a^2+2b^2+2c^2 \geq 2ab + 2bc + 2ac \]

Answer 5

i would find the rate at which (a^2 + b^2 + c^2) incraeses

and find the rate that (a+b)^2+(b+c)^2+(a+c)^2 increases

if you know the first term increases faster, it will always be positive or >= 0

Answer 6

\[\Large (a-b)^2+(b-c)^2+(c-a)^2 \geq 0  \]

Answer 7

\[\Large a^2-2ab + b^2 + b^2 -2bc + c^2 + c^2 -2ac + a^2 \geq 0 \]

Answer 8

So as in the expression above.

Answer 9

The algebraic manipulation is a nightmare though, took me several tries too, I think I have seen something similar to this once a while ago, hence I remembered to add up the terms.

Answer 10

Thanks Spacelimbus.

Answer 11

You're welcome

Answer 12

How did you get the idea of do that?

Answer 13

See your original equation. Not the entire one, there it says.
\[\Large (a+b)^2+(b+c)^2+(a+c)^2 = 2a^2 +2b^2 +2c^2 + 2ab + 2bc + 2ac \]

Answer 14

After a while I thought that this is the same as we have in the step where you can say: \[\Large a^2+b^2+c^2 \geq ab + bc + ca \]

Answer 15

So it's basically the same expression (after multiplication with 2) just that after algebraic manipulation the ab, bc, ca, terms will be negative, hence the minus sign in the binomials.