• Oct 12th 2007, 02:39 PM
rlarach
Prove that the inequality
3a4 - 4a3b + b4 ≥ 0
holds for all real numbers a and b.

THANK YOU VERY MUCH FOR YOUR HELP!!!:)
• Oct 12th 2007, 03:33 PM
Soroban
Hello, rlarach!

Quote:

Prove that: .$\displaystyle 3a^4 - 4a^3b + b^4 \;\geq \;0$ for all real numbers $\displaystyle a$ and $\displaystyle b$

Let $\displaystyle f \:=\:3a^4-4a^3b + b^4$

I used the Factor Theorem a few times.

If $\displaystyle b = a$, the polynomial has value 0.
. . Hence, $\displaystyle (a -b)$ is a factor of $\displaystyle f.$

Using long division: .$\displaystyle 3a^3 -4a^3b + b^4\;=\;(a - b)(3a^3 - a^2b - ab^2 - b^3)$

And we find that $\displaystyle (a-b)$ is a factor of the cubic.

Long division: .$\displaystyle f \;=\;(a-b)(a-b)(3a^2+2ab+b^2)$

. . Hence, we have: .$\displaystyle f \;=\;(a-b)^2\left(2a^2 + [a+b]^2\right)$

We see that: $\displaystyle (a-b)^2 \:\geq\:0$

. . and that: .$\displaystyle 2a^2 + (a+b)^2\:\geq\;0$

Therefore, $\displaystyle f \:\geq\:0$ for all real $\displaystyle a,\,b$