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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: An interesting result
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[Post New]23 Apr 2008 20:13:00 IST
    Subject: An interesting result
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hsbhatt (3156)

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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
    
[Post New]24 Apr 2008 19:41:38 IST
    Subject: Re:An interesting result
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sboosy (2970)

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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
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[Post New]25 Apr 2008 14:24:26 IST
    Subject: Re:An interesting result
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hsbhatt (3156)

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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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[Post New]25 Apr 2008 21:03:27 IST
    Subject: Re:An interesting result
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oneyeartogo (188)

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