that's the answer
can u pls explain??
Let l,m,n be three consecutive natural numbers (l>m>n).If the angle A of \DeltaABC be given by
sinA=\frac{\left(a^{2}+b^{2}+c^{2} \right)\left(l^{2}+m^{2}+n^{2} \right)}\left(al+bm+cn)^{2}
and the perimeter of the triangle is 12,then the area of the triangle is??
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8 Answers
{\left(a^2+b^2+c^2\right)\left(l^2+m^2+n^2\right)}\left \geq (al+bm+cn)^2
thus we have that equality holds... 9because this is cauchy swarz.. which holds) when a/l=b/m=c/n
now you can find it?
sides: 3,4,5
area=6
In any case, by Cauchy Schwarz Inequality, we have
(al + bm + cn )2 ≤ (a2+ b2 + c2 )(l2 + m2 + n2 )
1 ≤ (a2+ b2 + c2 )(l2 + m2 + n2 )/(al + bm + cn )2
1 ≤ sinA
so sin A is 1 .. only equality is possible
A = 90°
perimeter is 12 so the sides are 3,4,5 ... area 6 !
riddle you missed one small point
equality holds when a/l=b/m=c/n
from there you get that the ratio of the sides is x:x+1:x+2
and hence that the sides are 3,4,5
chinmay there is a post here.. let me search for it
for IIT JEE
only AM GM is in syllabus.. But knowing some of these might help!
specially Cauchy schwarz
it says
|a.b|<|a||b| where. is the dot product!
equality holds when the vectors are parallel...!