Question

a=3,b=2,c=1 then a+b+c=?​

Answer 1

Answer:

Arithmetic Mean of n numbers will always be greater than equal to the Geometric Mean of the same n numbers”.

This sum is actually an application of the inequality. See how!!!

Let a-3, b-2 and c+1 be termed as x,y,z respectively.

Then by the inequality, we know that

(x+y+z)3≥(xyz)13

=>[(a−3)+(b−2)+(c+1)]3≥[(a−3)(b−2)(c+1)]13

=>(a+b+c−4)3￱≥[(a−3)(b−2)(c+1)]13

=>(13−4)3￱≥[(a−3)(b−2)(c+1)]13

=>3≥[(a−3)(b−2)(c+1)]13

=>(3)3≥(a−3)(b−2)(c+1)

=>(a−3)(b−2)(c+1)≤27

Thus the maximum value of (a-3)(b-2)(c+1) is 27.Arithmetic Mean of n numbers will always be greater than equal to the Geometric Mean of the same n numbers”.

This sum is actually an application of the inequality. See how!!!

Let a-3, b-2 and c+1 be termed as x,y,z respectively.

Then by the inequality, we know that

(x+y+z)3≥(xyz)13

=>[(a−3)+(b−2)+(c+1)]3≥[(a−3)(b−2)(c+1)]13

=>(a+b+c−4)3￱≥[(a−3)(b−2)(c+1)]13

=>(13−4)3￱≥[(a−3)(b−2)(c+1)]13

=>3≥[(a−3)(b−2)(c+1)]13

=>(3)3≥(a−3)(b−2)(c+1)

=>(a−3)(b−2)(c+1)≤27

Thus the maximum value of (a-3)(b-2)(c+1) is 27.Arithmetic Mean of n numbers will always be greater than equal to the Geometric Mean of the same n numbers”.

This sum is actually an application of the inequality. See how!!!

Let a-3, b-2 and c+1 be termed as x,y,z respectively.

Then by the inequality, we know that

(x+y+z)3≥(xyz)13

=>[(a−3)+(b−2)+(c+1)]3≥[(a−3)(b−2)(c+1)]13

=>(a+b+c−4)3￱≥[(a−3)(b−2)(c+1)]13

=>(13−4)3￱≥[(a−3)(b−2)(c+1)]13

=>3≥[(a−3)(b−2)(c+1)]13

=>(3)3≥(a−3)(b−2)(c+1)

=>(a−3)(b−2)(c+1)≤27

Thus the maximum value of (a-3)(b-2)(c+1) is 27.Arithmetic Mean of n numbers will always be greater than equal to the Geometric Mean of the same n numbers”.

This sum is actually an application of the inequality. See how!!!

Let a-3, b-2 and c+1 be termed as x,y,z respectively.

Then by the inequality, we know that

(x+y+z)3≥(xyz)13

=>[(a−3)+(b−2)+(c+1)]3≥[(a−3)(b−2)(c+1)]13

=>(a+b+c−4)3￱≥[(a−3)(b−2)(c+1)]13

=>(13−4)3￱≥[(a−3)(b−2)(c+1)]13

=>3≥[(a−3)(b−2)(c+1)]13

=>(3)3≥(a−3)(b−2)(c+1)

=>(a−3)(b−2)(c+1)≤27

Thus the maximum value of (a-3)(b-2)(c+1) is 27.