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 $\displaystyle a=b=c$.
How about this 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.