Question
Question: If \[{a_1},{a_2}, \ldots \ldots ,{a_n}\] are positive real numbers whose product is a fixed number \...
If a1,a2,……,an are positive real numbers whose product is a fixed number c, then the minimum value of a1+a2+……+an−1+2an is
(a) n(2c)1/n
(b) (n+1)c1/n
(c) 2nc1/n
(d) (n+1)(2c)1/n
Solution
Here, we will first find the arithmetic mean and geometric mean of the terms in the expression. Then, we will use the relation between arithmetic mean and geometric mean to form an inequation. Finally, we will use the given information to find the minimum value of the expression a1+a2+……+an−1+2an.
Formula Used:
We will use the following formulas:
The arithmetic mean of the n numbers a1,a2,……,an is given by the formula A.M.=na1+a2+……+an−1+an.
The geometric mean of the n numbers a1,a2,……,an is given by the formula G.M.=na1a2……an−1an.
Complete step-by-step answer:
We will use the formula for A.M. and G.M. to find the minimum value of a1+a2+……+an−1+2an.
The number of terms in the sum a1+a2+……+an−1+2an is n.
Therefore, using the formula A.M.=na1+a2+……+an−1+an, we get the arithmetic mean as
A.M.=na1+a2+……+an−1+2an
The number of terms in the sum a1+a2+……+an−1+2an is n.
Therefore, using the formula G.M.=na1a2……an−1an, we get the geometric mean as
G.M.=(a1a2……an−12an)1/n
Now, we know that the arithmetic mean is always greater than or equal to the geometric mean.
Therefore, we get
⇒A.M.≥G.M.
Substituting A.M.=na1+a2+……+an−1+2an and G.M.=(a1a2……an−12an)1/n in the inequation, we get
⇒na1+a2+……+an−1+2an≥(a1a2……an−12an)1/n
Rewriting the inequation, we get
⇒na1+a2+……+an−1+2an≥(2a1a2……an−1an)1/n
It is given that the number a1,a2,……,an are positive real numbers whose product is a fixed number c.
Therefore, we get
a1a2……an−1an=c
Substituting a1a2……an−1an=c in the inequation na1+a2+……+an−1+2an≥(2a1a2……an−1an)1/n, we get
⇒na1+a2+……+an−1+2an≥(2c)1/n
Multiplying both sides by n, we get
⇒n(na1+a2+……+an−1+2an)≥n(2c)1/n
Thus, we get
⇒a1+a2+……+an−1+2an≥n(2c)1/n
Therefore, the value of the expression a1+a2+……+an−1+2an is greater than or equal to n(2c)1/n.
Thus, the minimum value of the expression a1+a2+……+an−1+2an is n(2c)1/n.
The correct option is option (a).
Note: We multiplied both sides of the inequation na1+a2+……+an−1+2an≥(2c)1/n by n. Since the number of terms cannot be negative, n is a positive integer. Therefore, we could multiply both sides of the inequation by n without changing the sign of the inequation.
Here we used geometric mean and arithmetic mean to solve the question. These are the two types of mean and the third type of mean is harmonic mean.