Inequality. (1 Viewer)

[conics2008]

just a quick one,..

given a^2+b^2 > 2ab and that a^2+b^2+c^2 > ab+ac+bc €

prove using the above statments that

a^2b^2 +b^2c^2+c^2a^2 /a+b+c > abc

thanks for your time.

 

[3.14159potato26]

just a quick one,..

given a^2+b^2 > 2ab and that a^2+b^2+c^2 > ab+ac+bc €

prove using the above statments that

a^2b^2 +b^2c^2+c^2a^2 /a+b+c > abc

Proof:
If a^2+b^2+c^2 > ab+ac+bc,
then (ab)^2+(bc)^2+(ca)^2 > ab^2c + bc^2a + ca^2b
(ab)^2+(bc)^2+(ca)^2 > abc(a+b+c)
(a^2b^2 +b^2c^2+c^2a^2) / (a+b+c) > abc

 

[conics2008]

huh ^^^^^^^^^

I didn't get a single line of your working out, can you please explain.

 

[shinn]

SINCE A^2+B^2+C^2 > AB+AC+BC,

Let:
A = xy
B = xz
C = yz

Therefore,
x^2y^2 + x^2z^2 + y^2z^2
> x^2yz + y^2xz + z^2xy
= xyz (x+y+z)

Dividing both sides by x+y+z and Reassigning x = a, y= b, z= c:

(a^2b^2 +b^2c^2+c^2a^2) / (a+b+c) > abc

 

[friction]

Squareds in the middle of terms make is so confusing.
Ill explain potatoes.


Proof:
If a^2+b^2+c^2 > ab+ac+bc,
a=ab
b=bc
c=ac

then (ab)^2+(bc)^2+(ca)^2 > ac(b^2) + ba(c^2) + cb(a^2)

Take abc out of the RHS

(ab)^2+(bc)^2+(ca)^2 > abc(a+b+c)

take a + b + c over the other side.

(a^2b^2 +b^2c^2+c^2a^2) / (a+b+c) > abc

 

[undalay]

nvm read question wrong

edit: Frictions working seems correct