IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

$\text {Let the roots be} \ -a_1, -a_2, ..,-a_n \\ \\<br/>\text{where} \ a_i \ge 0 \ \text{for} \ 1 \le i \le n$

Then f(x) = (x+a_1) (x+a_2) (x+a_3)...(x+a_n)

We can see that a_1 a_2...a_n = 1

Each coefficient is a sum of products of a_i ' taken r at a time.

Consider say $a_1 a_2 a_3...a_r + a_2 a_3...a_{r+1} + ...+ a_{n-r+1} a_{n-r+2}...a_n$ which is the coefficient of $x^{n-r}$ . The number of terms is $\binom {n} {r}$

By AM-GM Inequality $\frac{\sum a_1 a_2...a_r} {\binom {n}{r}} \ge 1$

Hence $\sum a_1 a_2...a_r \ge \binom {n}{r}$

Thus $f(x) \ge x^n + \binom {n}{1} x^{n-1} + \binom {n}{2} x^{n-2} + ...+ \binom {n}{n-1} x + 1 = (1+x)^n$ as sboosy pointed out.

Hence $f(2) \ge (1+2)^n = 3^n$

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