# Another proof

• Jul 5th 2006, 04:04 PM
Nichelle14
Another proof
I have to Prove (a+b)^2 + (b+c)^2 + (c+a)^2 >= 4(ab + bc + ac)

I multiplied out both sides and then move everything to the left, but I don't know what I should do after that.
• Jul 5th 2006, 06:32 PM
malaygoel
Quote:

Originally Posted by Nichelle14
I have to Prove (a+b)^2 + (b+c)^2 + (c+a)^2 >= 4(ab + bc + ac)

I multiplied out both sides and then move everything to the left, but I don't know what I should do after that.

$(a+b)^2 + (b+c)^2 + (c+a)^2$
$=2(a^2 + b^2 + c^2)+ 2(ab + bc + ca)$
$=(a^2 + b^2)+(b^2 + c^2)+(a^2 + c^2)+ 2(ab + bc + ca)$
$\geq4(ab + bc + ca)$

Keep Smiling
Malay
• Jul 5th 2006, 06:54 PM
ThePerfectHacker
Quote:

Originally Posted by Nichelle14
I have to Prove (a+b)^2 + (b+c)^2 + (c+a)^2 >= 4(ab + bc + ac)

I multiplied out both sides and then move everything to the left, but I don't know what I should do after that.

It is necessary and suffienct for,
$a^2+b^2+c^2\geq ab+ac+bc$
----
If you play your cards right this is the "Cauchy-Swarchtz Inequality".
Let,
$\bold{v}=(a,b,c)$
$\bold{u}=(b,c,a)$
Then,
$||\bold{u}||\cdot ||\bold{v}||\geq |\bold{u}\cdot \bold{v} |$
Now,
$
||\bold{u}||\cdot ||\bold{v}||=a^2+b^2+c^2$

And,
$|\bold{u}\cdot \bold{v} |=|ab+bc+ac|$
Thus,
$a^2+b^2+c^2\geq |ab+bc+ac|\geq ab+bc+ac$
• Jul 6th 2006, 09:18 AM
Soroban
Hello, Nichelle14!

Quote:

Prove that: . $(a+b)^2 + (b+c)^2 + (c+a)^2\:\geq\:4(ab + bc + ac)$

I multiplied out both sides and then move everything to the left. . . . Not recommended

Multiply out the left side and we have:
. . $(a+b)^2 + (b+c)^2 + (c+a)^2\:=$ $\:2(a^2 + b^2+c^2) + 4(ab + bc + ac)$ LHS

Since $a,\,b,\,c$ are real numbers: . $a^2 + b^2 + c^2 \:\geq \:0$

Multiply by $2:\;\;2(a^2 + b^2 + c^2) \:\geq \:0$

Add $4(ab + bc + ac)$ to both sides:

. . $\underbrace{2(a^2 + b^2 + c^2) + 4(ab + bc + ac)} \:\geq\:4(ab + bc + ac)$
. . . . . . . . This is LHS

Therefore: . $(a+b)^2 + (b+c)^2 + (c+a)^2\;\geq\;4(ab + bc + ac)$