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Algebra

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23 Apr 2008 20:13:00 IST

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Consider the polynomial:

$f(x) = x^n + a_1 x^{n-1} + a_2 x^{n-2} + ...+ a_{n-1} x + 1$

It is given that it has n real roots and $a_i \ge 0, 1\le i \le n-1$

Prove that $f(2) \ge 3^n$

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Comments (3)

sreeraman nagasubramaniyan

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24 Apr 2008 19:41:38 IST

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$f(x)=(1+x)^2 = x^2+2x+1 , f(x)=0 \ \mbox{has repeated root} \ -1 \\ \\ \mbox{Also} \ x^2+3x+1=0,x^2+4x+1=0 ...,x^2+ax+1 (a\geq2) \\ \\ \mbox{always have real roots.Also} \ x^2+ax+1(a\geq2)\geq x^2+2x+1 = (1+x)^2(x \ \mbox{positive}) \\ \\ \mbox{Now just extending this concept to n from power 2} \\ \\ x^n+a_1x^{n-1}+a_2x^{n-2}+....1 \geq x^n+nC_1x^{n-1}+nC_2x^{n-2}+....1 = (1+x)^n \\ \\ \mbox{Thus} \ f(2)\geq (1+2)^n = 3^n$

Hari Shankar

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25 Apr 2008 14:24:26 IST

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

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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Joined: 14 Apr 2008

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25 Apr 2008 21:03:27 IST

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wow !!!!!!! What a solution !!!! (I am absolutely stunned)

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