## Friday, November 25, 2005

### Mix it Up. Topic: Polynomials/Inequalities. Level: Olympiad.

Problem #1: (ACoPS 5.5.22) Let $P$ be a polynomial with positive coefficients. Prove that if

$P\left(\frac{1}{x}\right) \ge \frac{1}{P(x)}$

holds for $x=1$ then it holds for every $x > 0$.

Solution: Let $P(x) = a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$.

We have $P(1) \ge \frac{1}{P(1)} \Rightarrow [P(1)]^2 = (a_n+a_{n-1}+\cdots+a_0)^2 \ge 1$.

Then we have $P(x) \cdot P\left(\frac{1}{x}\right) = (a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0)\left(a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+ \cdots + a_0)$.

But $(a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0)\left(a_n\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+ \cdots + a_0) \ge (a_n+a_{n-1}+\cdots+a_0)^2 \ge 1$

by Cauchy (see Post 12: November 24th, 2005). QED.

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Problem #2: (USAMO 1983 #2) Prove that the zeros of

$x^5+ax^4+bx^3+cx^2+dx+e = 0$

cannot all be real if $2a^2 < 5b$.

Solution: Hmm... coefficients of polynomials... let's break out Vieta's Formulas !

So we have $a = -(r_1+r_2+r_3+r_4+r_5) = -\sum r_i$ (for shorthand).

And $b = r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5 = \sum r_ir_j$ (again, for shorthand).

So we have $2a^2<5b \Rightarrow 2\left(\sum r_i\right)^2 < 5\sum r_ir_j$.

Expanding and simplifying, we have $2\left(\sum r_i^2\right) < \sum r_ir_j$ which can be rearranged to

$\sum (r_i-r_j)^2 < 0$ which clearly cannot hold if all the roots are real. QED.

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Practice Problem #1: Given the polynomial $x^3+ax^2+bx+c$ with real coefficients, find the condition the polynomial must satisfy such that it has at least one nonreal root (based on $a,b,c$). And generalize for $a_nx^n+a_{n-1}x^{n-1}+ \cdots + a_0$.

Practice Problem #2: (ACoPS 5.5.36) Prove Cauchy-Schwarz using the polynomial

$f(x) = (a_1x+b_1)^2+(a_2x+b_2)^2+\cdots+(a_nx+b_n)^2$

by observing that $f$ has real zeros iff $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}$.

1. ......which can be rearranged to (image)

which clearly cannot hold if all the roots are real. QED.

Can you explain how you rearranged?

2. Nice solution to #1!

3. @1234567890: Put all the terms on the LHS, multiply by 2, and rearrange things so you get something like this - a^2-2ab+b^2 = (a-b)^2 for each pair of the roots.

4. [...] We remember back here we used the one and only Vieta’s Formulas, and we can do so again. [...]