Question

suppose the sum of the seven positive numbers is 21.what is the minimum possible value of the average of the square of these numbers?​

Answer 1

Step-by-step explanation:

a+b+c+d+e+f+g = 21

for a²,b²,c²,d²,e²,f²,g²

AM ≥ GM

( a² + b² + c² + d² + e² + f² + g² ) / 7 ≥ (a²b²c²d²e²f²g² ) ^ (1/7)

( a² + b² + c² + d² + e² + f² + g² ) / 7 ≥ [ (abcdefg ) ^ (1/7) ]²

For abcdefg

AM ≥ GM

( a + b + c + d + e + f + g ) / 7 ≥ (abcdefg)^(1/7)

3 ≥ (abcdefg)^1/7

( a² + b² + c² + d² + e² + f² + g² ) ≥ 7 * [ (abcdefg ) ^ (1/7) ]²

hence minimum possible value for it is 7 * [ (abcdefg ) ^ (1/7) ]²

but since maximum value possible for (abcdefg)^1/7 is 3

we put it 3 in 7 * [ (abcdefg ) ^ (1/7) ]²

7 * 3²

7*9

63 is your answer

please mark as brainliest answer