# Homework Help: Algebra (Polynomials)

1. Jul 13, 2013

### Prashant Jain

Prove that product of sum of roots and sum of reciprocal of roots of a polynomial with degree n is always greater than or equal to n2.

I tried the same on a polynomial of degree 4:

ax4+bx3+cx2+dx+e = 0

Let the roots be p, q, r, and s

The following equations show the relation of roots to the coefficients of the polynomial

p + q + r + s = -b/a

pq + qr + rs + sp + pr + qs = c/a

pqr + qrs + rsp + spq = -d/a

pqrs = e/a

A general solution to the problem would be preferred.....

Last edited: Jul 13, 2013
2. Jul 13, 2013

### Infrared

Notice that if all of the roots are one, then the product of the sum of the roots and the sum of the reciprocals of the roots is $n^2$. So if you can prove that the case of all ones is a minimum, then you are done. One way to show this would be to look at $$(r_1+r_2+r_3+...r_n)(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...\frac{1}{r_n})$$ and see what would happen if we replace one of the r's with 1. You can show that this will always decrease the product by noting that $$p+\frac{1}{p} \geq 2$$ where p is positive (and the inequality is strict is p is not one).

3. Jul 13, 2013

### Prashant Jain

Sir, could the solution be a little more general?...... i.e. to help me prove this property is universally applicable for all the polynomials with real coefficients.
How to explain that the case where all roots are equal to one is a minimum?

4. Jul 13, 2013

### Infrared

How is my solution not general? It would prove it for all real polynomials. I suppose that I assumed that the roots were all positive, which you did not explicitly state, but this is a very necessary condition! Consider the quadratic $x^2-1$. Its roots are $\pm 1$ so the product you are considering would be zero, which is not greater than 2^2=4

How to show that the all ones case is the minimum? Do as I suggested in my previous post and replace just one of the roots with 1 and show that this reduces the product (using the inequality I gave you. You should prove this inequality if you have not seen it before). Once you show this, you are done because this implies

$$n^2=(1+1+1+...1)(1+1+1+...1) \leq (r_1+1+1+...1)(\frac{1}{r_1}+1+1+...1) \leq (r_1+r_2+1+...1)(\frac{1}{r_1}+\frac{1}{r_2}+1+...1)... \\ \leq (r_1+r_2+r_3+...r_n)(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+...\frac{1}{r_n})$$ as desired.

Last edited: Jul 13, 2013