SOLUTION TO PROBLEM 12

Solution # 1:
Consider the polynomial, p(x)=3x⁴-4x³b+b⁴. Since p(b)=0, bis a root of the polynomial and so (x-b) is a factor. Perform the long division to find that $p(x)=(x-b)\cdot(3x^3-x^2b-xb^2-b^3)$. Now consider q(x)=3x³-x²b-xb²-b³. Again b is a root and so (x-b) is a factor. Again perform the long division to find that $q(x)=(x-b)\cdot (3x^2+2xb+b^2)=(x-b)\cdot(2x^2+(x+b)^2$. Therefore $p(x)=(x-b)^2\cdot(2x^2+(x+b)^2$. Since $p(a)=(a-b)^2\cdot(2a^2+(a+b)^2)$ and $(a-b)^2\cdot(2a^2+(a+b)^2)\geq 0$, we have proven that $p(a)\geq 0$.

Solution # 2:
Use the Arithmetic-Geometric Mean Inequality,

$$\frac{a^4+a^4+a^4+b^4}{4}\geq \sqrt[4]{a^{12}b^4}=a^3b.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder
 

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Last modified:   Mon Apr 2 20:06:39 CDT 2001