1. ## inequality problem

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

2. Originally Posted by shosho
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, $\displaystyle a,b,c>0$:
$\displaystyle a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$
Because most of these inequalities are cyclic.

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

By the Cauchy-Swartz Inequality again:
$\displaystyle a^2b^2+b^2c^2+a^2c^2 = (ab)^2+(bc)^2+(ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)=$$\displaystyle a^2bc+b^2ac+c^2ab Thus, (by the transitive inequality ) \displaystyle a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab And only equality when \displaystyle a=b=c. 3. thanx for the help but im not sure about it... i dont really understand the theorem thing you used. could you please explain in another way it would be much appreciated 4. Originally Posted by shosho thanx for the help but im not sure about it... i dont really understand the theorem thing you used. could you please explain in another way it would be much appreciated Tell us what course this is part off, or what you were covering imeadiatly before the question was set. RonL 5. Originally Posted by ThePerfectHacker I think you means, \displaystyle a,b,c>0: \displaystyle a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab Because most of these inequalities are cyclic. Now, by the Cauchy-Swartz Inequality: \displaystyle a^4+b^4+c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \geq a^2b^2+b^2c^2+a^2c^2 By the Cauchy-Swartz Inequality again: \displaystyle a^2b^2+b^2c^2+a^2c^2 = (ab)^2+(bc)^2+(ac)^2 \geq (ab)(ac)+(bc)(ab)+(ac)(bc)=$$\displaystyle a^2bc+b^2ac+c^2ab$

Thus, (by the transitive inequality )
$\displaystyle a^4+b^4+c^4 \geq a^2bc+b^2ac+c^2ab$
And only equality when $\displaystyle a=b=c$.
The restriction $\displaystyle a,b,c>0$ is not needed.

RonL

6. 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

Thank you

7. Originally Posted by shosho
could you please explain in another way

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

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

Which is the Cauchy-Swartz inequality.