inequality problem

shosho · #1 $Old$ June 21st, 2007, 04:42 AM

prove that

a^4 + b^4 + c^2 greater than or equal to a^2bc + b^2ac + c^2ab

a,b,c are all real numbers

thanx for any help

ThePerfectHacker · #2 $Old$ June 21st, 2007, 09:11 PM

Quote:

Originally Posted by shosho $View Post$

prove that

a^4 + b^4 + c^2 greater than or equal to a^2bc + b^2ac + c^2ab

a,b,c are all real numbers

thanx for any help

I think you means, $a,b,c>0$ :
$a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$
Because most of these inequalities are cyclic.

Now, by the Cauchy-Swartz Inequality:
$a^4+b^4+c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \geq a^2b^2+b^2c^2+a^2c^2$

By the Cauchy-Swartz Inequality again:
$a^2b^2+b^2c^2+a^2c^2 = (ab)^2+(bc)^2+(ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)=$ $a^2bc+b^2ac+c^2ab$

Thus, (by the transitive inequality

)
$a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$
And only equality when $a=b=c$ .

shosho · #3 $Old$ June 25th, 2007, 01:52 AM

thanx for the help

but im not sure about it...
i dont really understand the theorem thing you used.

could you please explain in another way

it would be much appreciated

CaptainBlack · #4 $Old$ June 25th, 2007, 04:12 AM

Quote:

Originally Posted by shosho $View Post$

thanx for the help

but im not sure about it...
i dont really understand the theorem thing you used.

could you please explain in another way

it would be much appreciated

Tell us what course this is part off, or what you were covering imeadiatly
before the question was set.

RonL

CaptainBlack · #5 $Old$ June 25th, 2007, 04:13 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

I think you means, $a,b,c>0$ :
$a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$
Because most of these inequalities are cyclic.

Now, by the Cauchy-Swartz Inequality:
$a^4+b^4+c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \geq a^2b^2+b^2c^2+a^2c^2$

By the Cauchy-Swartz Inequality again:
$a^2b^2+b^2c^2+a^2c^2 = (ab)^2+(bc)^2+(ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)=$ $a^2bc+b^2ac+c^2ab$

Thus, (by the transitive inequality

)
$a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$
And only equality when $a=b=c$ .

The restriction $a,b,c>0$ is not needed.

RonL

shosho · #6 $Old$ June 25th, 2007, 07:06 AM

the course i was previously doiing was inequality. it wasnt all that of a difficult section of inequality. nothing specific but the question was given as an extension type. i seem to be terribly stuck
please help .. this is urgent

Thank you

ThePerfectHacker · #7 $Old$ June 25th, 2007, 03:59 PM

Quote:

Originally Posted by shosho $View Post$

could you please explain in another way

How about this way?

We need to use the inequality:
$x^2+y^2+z^2 \geq xy+yz+xz$
For all real numbers $x,y,z>0$ .

Begin by noticing that, by the AM-GM inequality,
$x^3+y^3+z^3 \geq 3xyz$
Thus,
$x^3+y^3+z^3 - 3xyz \geq 0$
Factor,
$(x+y+z)(x^2+y^2+z^2 - xy - yz - xz) \geq 0$
Since $(x+y+z)>0$ we can cancel to obtain,
$x^2+y^2+z^2 \geq xy+yz+xz$

Which is the Cauchy-Swartz inequality.