*Posted by Albert in : potd , trackback
*
Given the following polynomial with real coefficients and positive real roots:

$$f(x) = x^6-a x^5+b x^4-c x^3+d x^2-e x+64$$

What is the minimum possible value for $e$?

show

**Solution:**
We know $f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)(x-r_6)$, where $r_k$ is the $k$th root. This means the term with no $x$’s can only be obtained by multiplying (the negative of) all the roots together (this is a well known result, as per Vieta’s Formulas). Then $64 = r_1 r_2 r_3 r_4 r_5 r_6$. By a similar argument, $e = r_1 r_2 r_3 r_4 r_5 + \cdots + r_2 r_3 r_4 r_5 r_6$.

It is somewhat obvious that $e$ is minimized when $r_1 = r_2 = r_3 = r_4 = r_5 = r_6$. With this knowledge, $64 = r_1^6 \Rightarrow r_1 = r_2 = r_3 = r_4 = r_5 = r_6 = 2$. Then, $e = 6(2^5) = \fbox{192}$.

For a more rigorous proof that this is correct, we can consider the GM-HM inequality on $r_1$ through $r_6$:

$$\sqrt[6]{r_1 r_2 r_3 r_4 r_5 r_6} \geq \frac{6}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}+\frac{1}{r_5}+\frac{1}{r_6}}$$

$$2 \geq \frac{6}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}+\frac{1}{r_5}+\frac{1}{r_6}}$$

$$\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}+\frac{1}{r_5}+\frac{1}{r_6} \geq 3$$

$$\frac{r_1 r_2 r_3 r_4 r_5 + \cdots + r_2 r_3 r_4 r_5 r_6}{r_1 r_2 r_3 r_4 r_5 r_6} \geq 3$$

$$\frac{e}{64} \geq 3$$

$$e \geq 192$$

Thus, the answer is $\fbox{192}$.

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