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.
$\displaystyle (a+b)^2 + (b+c)^2 + (c+a)^2$Originally Posted by Nichelle14
$\displaystyle =2(a^2 + b^2 + c^2)+ 2(ab + bc + ca)$
$\displaystyle =(a^2 + b^2)+(b^2 + c^2)+(a^2 + c^2)+ 2(ab + bc + ca)$
$\displaystyle \geq4(ab + bc + ca)$
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Malay
It is necessary and suffienct for,Originally Posted by Nichelle14
$\displaystyle a^2+b^2+c^2\geq ab+ac+bc$
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If you play your cards right this is the "Cauchy-Swarchtz Inequality".
Let,
$\displaystyle \bold{v}=(a,b,c)$
$\displaystyle \bold{u}=(b,c,a)$
Then,
$\displaystyle ||\bold{u}||\cdot ||\bold{v}||\geq |\bold{u}\cdot \bold{v} |$
Now,
$\displaystyle
||\bold{u}||\cdot ||\bold{v}||=a^2+b^2+c^2$
And,
$\displaystyle |\bold{u}\cdot \bold{v} |=|ab+bc+ac| $
Thus,
$\displaystyle a^2+b^2+c^2\geq |ab+bc+ac|\geq ab+bc+ac$
Hello, Nichelle14!
Prove that: .$\displaystyle (a+b)^2 + (b+c)^2 + (c+a)^2\:\geq\:4(ab + bc + ac)$
I multiplied out both sides and then move everything to the left. . . . Not recommended
Multiply out the left side and we have:
. . $\displaystyle (a+b)^2 + (b+c)^2 + (c+a)^2\:=$ $\displaystyle \:2(a^2 + b^2+c^2) + 4(ab + bc + ac)$ LHS
Since $\displaystyle a,\,b,\,c$ are real numbers: .$\displaystyle a^2 + b^2 + c^2 \:\geq \:0$
Multiply by $\displaystyle 2:\;\;2(a^2 + b^2 + c^2) \:\geq \:0$
Add $\displaystyle 4(ab + bc + ac)$ to both sides:
. . $\displaystyle \underbrace{2(a^2 + b^2 + c^2) + 4(ab + bc + ac)} \:\geq\:4(ab + bc + ac)$
. . . . . . . . This is LHS
Therefore: .$\displaystyle (a+b)^2 + (b+c)^2 + (c+a)^2\;\geq\;4(ab + bc + ac)$