Looks like Harmonic Mean -Arithmetic Mean Inequality
(but some of a,b,c could be negative? unless you omitted that they must be positive [which you later confirmed]).
(Note: Until you said a,b,c positive, we can't just simply multiply across by the denominators (a+b+c)*(ab+bc+ca) and start reducing, since we don't know whether either/both/neither of those brackets is +ve/-ve.)
Either we can try to prove it directly, or we can start with the result to prove, then work it backwards to reduce it to the HM-AM Inequality or somesuch.
To prove: 1+ 3/(a+b+c)≧6/(ab+bc+ca)
The Harmonic Mean -Geometric Mean-Arithmetic Mean Inequality (which is a special case of the Power-Mean or Holder Inequality)
says that for any *positive* real p,q,r:
Arithmetic Mean >= their Harmonic Mean :
(p+q+r)/3 >= 3/(1/p + 1/q + 1/r)
but here we were not told that a,b,c were positive [later told this].
Arithmetic Mean >= Geometric Mean:
(p+q+r)/3 >= (pqr)^1/3
Geometric Mean >= Harmonic Mean:
(pqr)^1/3 >= 3/(1/p + 1/q + 1/r)
But in this specific case we know abc=1, thus (abc)^1/3 = 1
Thus the AM-GM-HM Inequality gives us:
(a+b+c)/3 >= 1 >= 3/(a+b+c)
Inverting the AM-GM Inequality and reversing inequality sign:
3/(a+b+c) <= 1
Finally this:
1+ 3/(a+b+c)≧6/(ab+bc+ca)
plugging in 1>= 3/(a+b+c)
1+ 3/(a+b+c) >= 3/(a+b+c) + 3/(a+b+c) = 6/(a+b+c)
So now to prove:
6/(a+b+c)≧6/(ab+bc+ca)
1/(a+b+c)≧1/(ab+bc+ca)
Invert both sides, swap inequality sign:
(a+b+c) <= (ab+bc+ca)
Divide RHS by abc=1
(a+b+c) <= (1/a + 1/b + 1/c)
This is much harder to prove, so maybe the step above plugging in 1>= 3/(a+b+c) was bad.
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* Bringing the RHS term to the LHS is no use:
1+ 3/(a+b+c)-6/(ab+bc+ca)≧0
* Here is a something that might give you a toehold on the problem:
HM-AM Inequality for p=bc, q=ac, r=ac (if a,b,c>0 then p,q,r>0)
(ab+ac+bc)/3 ≧ 3/(1/bc + 1/ac + 1/ab)
Now multiply the RHS denominator by abc = 1
(ab+ac+bc)/3 ≧ 3/(abc/bc + abc/ac + abc/ab)
(ab+ac+bc)/3 ≧ 3/(a+b+c)
Invert both sides, swap inequality sign:
3/(ab+ac+bc) ≤ (a+b+c)/3
i.e. (a+b+c)/3 ≧ 3/(ab+ac+bc)
It seemed useful but the brackets are not both denominators as in the question.
2007-07-20 23:53:51
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answer #1
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answered by smci 7
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As your question is written, a, b, and c are all 1.
Since you didn't specify that a, b, and c were different numbers, 1 solves. 1 x 1 x 1 =1
1 + 3/(1+1+1)>=6/(1*1+1*1+1*1)
1 + 3/3 >=6/3
2 >= 2
2007-07-26 06:21:55
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answer #2
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answered by Anonymous
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everytime you multiply or divide via a damaging variety the direction of the inequality ameliorations working example -a million <= 10 if i divide via -a million the two aspects with out changing the sign direction see what occurs a million <= -10 this is obviously incorrect the actual answer is a million >= -10 so which you notice once you divide via negatives or multiply via them replace the direction of the better than below a million) -2*|2x+ 3| >= 4 after dividing via -2 |2x + 3| <= 4/(-2) |2x + 3| <= -2 now sparkling up it first we sparkling up this via noticing 2x + 3 <= -2 2x <= -2 -3 2x <= -5 x <= -5/2 [-5/2, -infinity) or -(2x + 3) <= -2 2x + 3 >= 2 2x >= 2 - 3 2x >= -a million x >= -0.5 (infinity , -a million/2] you employ the sq. brackets because of the fact meaning it could additionally be precisely -a million/2 =================== if it became x> -3/4 then you somewhat could write (infinity , -3/4) the open bracket skill it cant ever attain the cost of -3/4 yet can get clossseeee ========================== now the humorous factor is the solutions you eventually had have been maximum suitable however the type you ARRIVED AT THEM became incorrect... your x strategies did no longer correspond to what your very final solutions have been ... you had written x>= -5/2 and x <= -a million/2 this is what you arrived at and meaning -a million/2 >= x so this implies -a million/2 is the optimum value and finding on the different x >= -5/2 this implies the backside value is -5/2 so your solutions could have been [-a million/2, -5/2] those are incorrect solutions which you whould are turning out to be to be from the type you solved the question my solutions have been maximum suitable
2016-12-10 18:12:33
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answer #3
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answered by boven 4
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re check the question
you have missed out something
2007-07-21 00:07:16
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answer #4
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answered by kapilbansalagra 4
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out of a,b,c two of the numbers must be zero
2007-07-20 23:53:03
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answer #5
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answered by Emperor 3
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