Given that the real numbers a , b , c , d , and e satisfy the equations:
a + b + c + d + e = 8
a 2 + b 2 + c 2 + d 2 + e 2 = 1 6 ,
the difference in the maximum and minimum values of a can be written as B A where A and B are positive, coprime integers. Find the value of A + B .
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In your second to last line, where does the 4 come from in 4 ( 1 6 − a 2 ) and why is the first equation in that line useful if it doesn't even have an a in it.
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Using Schwartz inequality u can write. b as b.1 c as c.1. similarly....all...this add up to 1^2+ 1^2+1^2+1^2=4
Really nyc solution :)
Cauchy Schwarz inequality is valid only for positive reals. Although I myself find this surprising that the answers do come by using the inequality even if the variables are not given to be positive. Can anyone tell why it works?
It is intuitive that at maximum and minimum values of a the other variables are equal, (by symmetry) Plugging back in the equation we get a + 4 b = 8 a 2 + 4 b 2 = 1 6 Solving these two we get the two values of a as 16/5 and 0. Hence the answer 2 1
I like this user for his quirky approaches which most of the time end up being right!!! So LOL!!!
this is not a rigorous proof.
this is valid if u consider the symmetry of the problem and the common observation that minimas and maximas occur usually with some special features
like a square has the largest area among all rectangles with common perimeter, or like the range of projectile thrown from any point to any other is maximum when the time taken for that is unique,,
but yes it is not rigorious
U fkd the value of question bro but nice method
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A familiar problem from USAMO....
b + c + d + e = 8 - a
b^2 + c^2 + d^2 + e^2 = 16 - a^2
Using Cauchy-Schwarz Inequality:
(1 + 1 + 1 + 1)(b^2 + c^2 + d^2 + e^2) >= (b + c + d + e)^2 4(16 - a^2) >= (8 - a)^2 5a^2 - 16a <= 0 and so..
0 <= a <= 16/5. Hence, the answer.