Question

If $$a_1,a_2,\dots \dots ,a_n$$ are positive real numbers whose product is a fixed number $$c$$, then the minimum value of $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$ is
(a) $$n{\left(2c\right)}^{1/n}$$
(b) $$\left(n+1\right)c^{1/n}$$
(c) $$2n{c}^{1/n}$$
(d) $$\left(n+1\right){\left(2c\right)}^{1/n}$$

Answer 1

Hint: 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 $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$.

Formula Used:
 We will use the following formulas:
The arithmetic mean of the $$n$$ numbers $$a_1,a_2,\dots \dots ,a_n$$ is given by the formula $$A.M.={\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+a_n}{n}}$$.
The geometric mean of the $$n$$ numbers $$a_1,a_2,\dots \dots ,a_n$$ is given by the formula $$G.M.=\sqrt[n]{a_1{a}_2\dots \dots a_{n-1}{a}_n}$$.

Complete step-by-step answer:
We will use the formula for A.M. and G.M. to find the minimum value of $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$.
The number of terms in the sum $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$ is $$n$$.
Therefore, using the formula $$A.M.={\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+a_n}{n}}$$, we get the arithmetic mean as
$$A.M.={\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}$$
The number of terms in the sum $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$ is $$n$$.
Therefore, using the formula $$G.M.=\sqrt[n]{a_1{a}_2\dots \dots a_{n-1}{a}_n}$$, we get the geometric mean as
$$G.M.={\left(a_1{a}_2\dots \dots a_{n-1}2a_n\right)}^{1/n}$$
Now, we know that the arithmetic mean is always greater than or equal to the geometric mean.
Therefore, we get
$$\Rightarrow A.M.\ge G.M.$$
Substituting $$A.M.={\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}$$ and $$G.M.={\left(a_1{a}_2\dots \dots a_{n-1}2a_n\right)}^{1/n}$$ in the inequation, we get
$$\Rightarrow {\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}\ge {\left(a_1{a}_2\dots \dots a_{n-1}2a_n\right)}^{1/n}$$
Rewriting the inequation, we get
$$\Rightarrow {\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}\ge {\left(2a_1{a}_2\dots \dots a_{n-1}{a}_n\right)}^{1/n}$$
It is given that the number $$a_1,a_2,\dots \dots ,a_n$$ are positive real numbers whose product is a fixed number $$c$$.
Therefore, we get
$$a_1{a}_2\dots \dots a_{n-1}{a}_n=c$$
Substituting $$a_1{a}_2\dots \dots a_{n-1}{a}_n=c$$ in the inequation $${\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}\ge {\left(2a_1{a}_2\dots \dots a_{n-1}{a}_n\right)}^{1/n}$$, we get
$$\Rightarrow {\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}\ge {\left(2c\right)}^{1/n}$$
Multiplying both sides by $$n$$, we get
$$\Rightarrow n\left({\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}\right)\ge n{\left(2c\right)}^{1/n}$$
Thus, we get
$$\Rightarrow a_1+a_2+\dots \dots +a_{n-1}+2a_n\ge n{\left(2c\right)}^{1/n}$$
Therefore, the value of the expression $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$ is greater than or equal to $$n{\left(2c\right)}^{1/n}$$.
Thus, the minimum value of the expression $$a_1+a_2+\dots \dots +a_{n-1}+2a_n$$ is $$n{\left(2c\right)}^{1/n}$$.
The correct option is option (a).

Note: We multiplied both sides of the inequation $${\displaystyle \frac{a_1+a_2+\dots \dots +a_{n-1}+2a_n}{n}}\ge {\left(2c\right)}^{1/n}$$ 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.