11Cosider the most general function that shall help us over here - f(x)=asin^2x+bsinxcosx+csin^2x
Using the fact that the maximum value of asinx+bcosx is \sqrt{a^2+b^2} we have the maximum value of f(x) to be \frac{a+c}{2}+\frac{\sqrt{(c-a)^2+b^2}}{2}
Here we substitute the values of a,b,c accordingly to get a reqd expression of L.H.S as
\frac{ab+1}{2}+\frac{\sqrt{(ab-1)^2+(a+b)^2}}{2}
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.
1how did u arrive at the max value of f(x)
1
Amritansh Bharech
·2010-01-11 11:42:29
and how r u showin max value of LHS to be less than RHS
by expanding the LHS we get
ab+12 + √(a2+1)(b2+1)2
then wat
1
Amritansh Bharech
·2010-01-12 09:52:39
okay i got the LHS less than RHS proof
but how did u get the max value of f(x) = asin2x+bsinxcosx+ccos2x
did u do maxima minima or some other way
11
Devil
·2010-01-15 09:14:12
I did not use maxima-minima.
There's however an alternate (& better) soln.
Consider
k=\left\{(sin\gamma+acos\gamma)+(sin\gamma+bcos\gamma) \right\}^2
Obvious that k=\left\{2sin\gamma+(a+b)cos\gamma \right\}^2\le 4\left\{1+\left(\frac{a+b}{2} \right)^2 \right\}
L=\frac{k}{4}=\left(\frac{sin\gamma+acos\gamma}{2} \right)^2+\left(\frac{(sin\gamma+acos\gamma)(sin\gamma+bcos\gamma)}{2} \right)+\left(\frac{sin\gamma+bcos\gamma}{2} \right)^2
A.M-G.M gives \boxed{(sin\gamma+acos\gamma)(sin\gamma+bcos\gamma)\le L\le 1+\left(\frac{a+b}{2} \right)^2}
1
Amritansh Bharech
·2010-01-19 11:04:33
i know that the max value of a sinx + b cosx is √a2 + b2
but how did u get the max value of a sin2x + b sinx cosx + c cos2x
and i didnt get the alternate soln. u gave
L is not the A.M. of (sinγ + a cosγ)2 and (sinγ + b cosγ)2
and how did u get k ≤ 4{1 + (a + b2)2}
btw u dont write anythin here but upload images i guess
where do u write them
341To summarise what Soumik has done. First remember that
xy \le \frac{x^2+y^2}{4}
(\sin \gamma + a \cos \gamma)(\sin \gamma + b \cos \gamma) \le \left(\sin \gamma + \frac{(a+b)}{2} \cos \gamma \right)^2
We know that |p \cos \alpha + q \sin \alpha| \le \sqrt{p^2+q^2}
Hence we have
\left(\sin \gamma + \frac{(a+b)}{2} \cos \gamma \right)^2 \le 1 + \frac{(a+b)^2}{4}
and the inequality is proved
1
Amritansh Bharech
·2010-01-21 12:34:45
firstly how is xy≤x2 + y24
it shud be xy≤x2 + y22
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
341
Hari Shankar
·2010-01-21 17:33:59
Sry, that was a typo. It should be xy \le \frac{(x+y)^2}{4} as has been used in the following steps
341
Hari Shankar
·2010-01-21 17:40:53
a \sin^2 x + b \sin x \cos x + c \cos^2 x = \frac{1}{2} [ a(1-\cos 2x) + b \sin 2x + c (1+\cos 2x) ]
= \frac{1}{2} [(a+c) + b \sin 2x + (c-a) \cos 2x} \le \frac{1}{2} [(a+c) + \sqrt{b^2 + (c-a)^2]
.
1(x-y)2≥0 or (x+y)2-4xy≥0 so xy≤(x+y)2/4.....