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Home » Mathematics » High School Pre-Olympiad (Ages 14+)
 
Inequality
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triplebig
Riemann Hypothesis
Riemann Hypothesis


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#1
Inequality
a+b+c=5 and ab+bc+ac=3
If a + b + c = 5 and ab + bc + ac = 3

prove that - 1\leq c\leq \frac {13}{3}

EDIT: a,b,c\in\mathbb{R}
Last edited by triplebig on Thu Nov 05, 2009 4:59 am; edited 1 time in total 
PostPosted: Thu Nov 05, 2009 3:19 am  Back to top 
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aadil
Riemann Hypothesis
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#2
what are a,b ?can they be negative
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PostPosted: Thu Nov 05, 2009 4:57 am  Back to top 
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triplebig
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#3
a,b,c are real numbers.

PostPosted: Thu Nov 05, 2009 4:59 am  Back to top 
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darkmage2009
Hodge Conjecture
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#4
This is not a rigorous or formal proof. It's also somewhat incomplete. It involves calculus as well.
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We can let a,b,c to be roots of a monic polynomial f(x). By Vieta's Laws, f(x)=x^3-5x^2+3x-k, where k=abc. Looking at the graph of f, we notice that it has a relative maximum and a relative minimum, so we must restrict k such that the relative maximum is greater than or equal to 0 and the relative minimum is less than or equal to 0 to ensure we have 3 real roots.

We can find the x-coordinates of the maximum and minimum, which are not based on the value of k. We take the derivative of f(x) and obtain f'(x)=3x^2-10x+3=(3x-1)(x-3). We set f'(x) equal to zero to find that the maximum occurs at x=\frac{1}{3} and the minimum occurs at x=3. For f(\frac{1}{3})\geq0, k\geq-\frac{13}{27} and for f(3)\leq0, k\leq9.

When f(\frac{1}{3})=0, let a=\frac{1}{3} and b=\frac{1}{3}. (If c is one of them, then the inequality obviously holds since -1<\frac{1}{3}<\frac{13}{3}.) Then c must equal \frac{13}{3} by the conditions given. When f(3)=0, let a=3 and b=3. (If c is one of them, then the inequality obviously holds since -1<3<\frac{13}{3}.) Then c must equal -1 by the conditions given.

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PostPosted: Thu Nov 05, 2009 12:29 pm  Back to top 
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Brut3Forc3
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#5
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The first equation gives a+b=5-c, and the second gives ab+(a+b)c=3. Plugging in for a+b, we have ab+(5-c)c=3, or ab=c^2-5c+3.
We have that a,b are the roots of x^2-(5-c)x+(c^2-5c+3), which has discriminant (5-c)^2-4(c^2-5c+3)=-3c^2+10c+13=-(c+1)(3c-13). This discriminant must be nonnegative for a,b to be real, so we have -1\le c \le \frac{13}{3}, as desired.
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PostPosted: Thu Nov 05, 2009 8:51 pm  Back to top 
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kunny
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#6
Those solution are correct.

Remark:

\boxed{\boxed{a,\ b\in\mathbb{R}\Longleftrightarrow a+b,\ a-b\in\mathbb{R}}}
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PostPosted: Thu Nov 05, 2009 9:02 pm  Back to top 
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