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#1 Posted 10:18am 09-01-10 InequalityProve that [im]http://codecogs.izyba.com/gif.latex?%28%5Csin%20%5Cgamma%20+%20a%20%5Ccos%20%5Cgamma%29%28%5Csin%20%5Cgamma%20+%20b%20%5Ccos%20%5Cgamma%29%20%5Cle%201%20+%20%5Cleft%28%20%5Cfrac%7Ba+b%7D%7B2%7D%20%5Cright%29%5E2[/im] for reals a,b |
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#2 Posted 10:39pm 11-01-10 Re: InequalityCosider the most general function that shall help us over here - [im]http://codecogs.izyba.com/gif.latex?f%28x%29%3Dasin%5E2x+bsinxcosx+csin%5E2x[/im] Using the fact that the maximum value of asinx+bcosx is [im]http://codecogs.izyba.com/gif.latex?%5Csqrt%7Ba%5E2+b%5E2%7D[/im] we have the maximum value of f(x) to be [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Ba+c%7D%7B2%7D+%5Cfrac%7B%5Csqrt%7B%28c-a%29%5E2+b%5E2%7D%7D%7B2%7D[/im] Here we substitute the values of a,b,c accordingly to get a reqd expression of L.H.S as [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7Bab+1%7D%7B2%7D+%5Cfrac%7B%5Csqrt%7B%28ab-1%29%5E2+%28a+b%29%5E2%7D%7D%7B2%7D[/im] Proving this to be less than R.H.S is simple, 1 method (that I used), was expansion of both sides - posting which is needless. |
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#3 Posted 01:05am 12-01-10 Re: Inequalityhow did u arrive at the max value of f(x) |
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#4 Posted 01:12am 12-01-10 Re: Inequalityand how r u showin max value of LHS to be less than RHS by expanding the LHS we get [frac]ab+1[/]2[/frac] + [frac][sqrt](a[p]2[/p]+1)(b[p]2[/p]+1)[/sqrt][/]2[/frac] then wat |
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#5 Posted 11:22pm 12-01-10 Re: Inequalityokay i got the LHS less than RHS proof but how did u get the max value of f(x) = asin[p]2[/p]x+bsinxcosx+ccos[p]2[/p]x did u do maxima minima or some other way |
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#6 Posted 10:44pm 15-01-10 Re: InequalityI did not use maxima-minima. There's however an alternate (& better) soln. Consider [im]http://codecogs.izyba.com/gif.latex?k%3D%5Cleft%5C%7B%28sin%5Cgamma+acos%5Cgamma%29+%28sin%5Cgamma+bcos%5Cgamma%29%20%5Cright%5C%7D%5E2[/im] Obvious that [im]http://codecogs.izyba.com/gif.latex?k%3D%5Cleft%5C%7B2sin%5Cgamma+%28a+b%29cos%5Cgamma%20%5Cright%5C%7D%5E2%5Cle%204%5Cleft%5C%7B1+%5Cleft%28%5Cfrac%7Ba+b%7D%7B2%7D%20%5Cright%29%5E2%20%5Cright%5C%7D[/im] [im]http://codecogs.izyba.com/gif.latex?L%3D%5Cfrac%7Bk%7D%7B4%7D%3D%5Cleft%28%5Cfrac%7Bsin%5Cgamma+acos%5Cgamma%7D%7B2%7D%20%5Cright%29%5E2+%5Cleft%28%5Cfrac%7B%28sin%5Cgamma+acos%5Cgamma%29%28sin%5Cgamma+bcos%5Cgamma%29%7D%7B2%7D%20%5Cright%29+%5Cleft%28%5Cfrac%7Bsin%5Cgamma+bcos%5Cgamma%7D%7B2%7D%20%5Cright%29%5E2[/im] A.M-G.M gives [im]http://codecogs.izyba.com/gif.latex?%5Cboxed%7B%28sin%5Cgamma+acos%5Cgamma%29%28sin%5Cgamma+bcos%5Cgamma%29%5Cle%20L%5Cle%201+%5Cleft%28%5Cfrac%7Ba+b%7D%7B2%7D%20%5Cright%29%5E2%7D[/im] |
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#7 Posted 8:21pm 19-01-10 Re: InequalityYup that's perfect. |
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#8 Posted 00:34am 20-01-10 Re: Inequalityi know that the max value of a sinx + b cosx is [sqrt]a[p]2[/p] + b[p]2[/p][/sqrt] but how did u get the max value of a sin[p]2[/p]x + b sinx cosx + c cos[p]2[/p]x and i didnt get the alternate soln. u gave L is not the A.M. of (sinγ + a cosγ)[p]2[/p] and (sinγ + b cosγ)[p]2[/p] and how did u get k ≤ 4{1 + ([frac]a + b[/]2[/frac])[p]2[/p]} btw u dont write anythin here but upload images i guess where do u write them |
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#9 Posted 09:51am 20-01-10 Re: InequalityTo summarise what Soumik has done. First remember that [im]http://codecogs.izyba.com/gif.latex?xy%20%5Cle%20%5Cfrac%7Bx%5E2+y%5E2%7D%7B4%7D[/im] [im]http://codecogs.izyba.com/gif.latex?%28%5Csin%20%5Cgamma%20+%20a%20%5Ccos%20%5Cgamma%29%28%5Csin%20%5Cgamma%20+%20b%20%5Ccos%20%5Cgamma%29%20%5Cle%20%5Cleft%28%5Csin%20%5Cgamma%20+%20%5Cfrac%7B%28a+b%29%7D%7B2%7D%20%5Ccos%20%5Cgamma%20%5Cright%29%5E2[/im] We know that [im]http://codecogs.izyba.com/gif.latex?%7Cp%20%5Ccos%20%5Calpha%20+%20q%20%5Csin%20%5Calpha%7C%20%5Cle%20%5Csqrt%7Bp%5E2+q%5E2%7D[/im] Hence we have [im]http://codecogs.izyba.com/gif.latex?%5Cleft%28%5Csin%20%5Cgamma%20+%20%5Cfrac%7B%28a+b%29%7D%7B2%7D%20%5Ccos%20%5Cgamma%20%5Cright%29%5E2%20%5Cle%201%20+%20%5Cfrac%7B%28a+b%29%5E2%7D%7B4%7D[/im] and the inequality is proved |
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#10 Posted 02:04am 22-01-10 Re: Inequalityfirstly how is xy≤[frac]x[p]2 + y[p]2[/p][/p][/]4[/frac] it shud be xy≤[frac]x[p]2 + y[p]2[/p][/p][/]2[/frac] then again how did u got to the 2nd step its not clear to me and the ques i asked above hav not been answered plz see the ques. i hav raised and try to eplain them to me |
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#11 Posted 07:03am 22-01-10 Re: InequalitySry, that was a typo. It should be [im]http://codecogs.izyba.com/gif.latex?xy%20%5Cle%20%5Cfrac%7B%28x+y%29%5E2%7D%7B4%7D[/im] as has been used in the following steps |
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#12 Posted 07:10am 22-01-10 Re: Inequality[im]http://codecogs.izyba.com/gif.latex?a%20%5Csin%5E2%20x%20+%20b%20%5Csin%20x%20%5Ccos%20x%20+%20c%20%5Ccos%5E2%20x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%20a%281-%5Ccos%202x%29%20+%20b%20%5Csin%202x%20+%20c%20%281+%5Ccos%202x%29%20%5D[/im] [im]http://codecogs.izyba.com/gif.latex?%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5B%28a+c%29%20+%20b%20%5Csin%202x%20+%20%28c-a%29%20%5Ccos%202x%7D%20%5Cle%20%5Cfrac%7B1%7D%7B2%7D%20%5B%28a+c%29%20+%20%5Csqrt%7Bb%5E2%20+%20%28c-a%29%5E2%5D[/im] . |
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#13 Posted 5:46pm 22-02-10 Re: Inequality(x-y)[p]2[/p]≥0 or (x+y)[p]2[/p]-4xy≥0 so xy≤(x+y)[p]2[/p]/4.....
avra ghosh Edited on 5:47pm 22-02-10 |
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