Question

For a triangle with sides a b c

3(bc+ca+cb) <= (a+b+c)^2 <=4(bc +ca+ab)

Answer 1

and the Question would be?

Answer 2

Show that it is so.

Answer 3

http://books.google.com.pk/books?id=MWdMlAfAALYC&pg=SA16-PA9&lpg=SA16-PA9&dq=3(bc%2Bca%2Bcb)+%3C%3D+(a%2Bb%2Bc)^2+%3C%3D4(bc+%2Bca%2Bab)&source=bl&ots=JKluDu75PB&sig=7wkEvA4pNLbRRd2JqQKnIct0bgU&hl=en&ei=wNk7Tq7DBoia-wabsdWeAg&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBUQ6AEwAA#v=onepage&q=3(bc%2Bca%2Bcb)%20%3C%3D%20(a%2Bb%2Bc)%5E2%20%3C%3D4(bc%20%2Bca%2Bab)&f=false

Answer 4

[a+b+c]^2=a^2+b^2+c^2+2(ab+bc+ca)
hence to prove
(bc+ca+ab) <= a^2+b^2+c^2 <=2(bc +ca+ab)
it is quite clear that
(bc+ca+ab) <=2(bc +ca+ab)
now we need to prove that
a^2+b^2+c^2 lies btwn the two

Answer 5

Good start...

Answer 6

AM>=GM
4ab<=(a+b)^2
a^2+b^2>=2ab
similarly
b^2+c^2>=2bc
c^2+a^2>=2ca
adding the three we get
2[a^2+b^2+c^2]>=2(ab+bc+ca)
a^2+b^2+c^2] >= ab+bc+ca