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Question: If \({{a}_{1}},{{a}_{2}},..{{a}_{n}}\) are positive real numbers whose product is a fixed number c, ...

If a1,a2,..an{{a}_{1}},{{a}_{2}},..{{a}_{n}} are positive real numbers whose product is a fixed number c, then the minimum value of a1+a2+...an−1+2an{{a}_{1}}+{{a}_{2}}+...{{a}_{n-1}}+2{{a}_{n}}.
A. n(2c)1/nn{{\left( 2c \right)}^{1/n}}
B. (n+1)c1/n(n+1){{c}^{1/n}}
C. 2nc1/n2n{{c}^{1/n}}
D. (n+1)(2c)1/n(n+1){{(2c)}^{1/n}}

Explanation

Solution

The given problem statement is very simple: you just need to apply basic logic with that you also need to apply the concept of the arithmetic mean and geometric mean. Now, you can simply solve the problem statement. So, let’s see what will be the approach for the given problem statement.

Step-By-Step Solution:
The given problem statement is to find the minimum value ofa1+a2+...an−1+2an{{a}_{1}}+{{a}_{2}}+...{{a}_{n-1}}+2{{a}_{n}}, when a1,a2,..an{{a}_{1}},{{a}_{2}},..{{a}_{n}}are positive real numbers whose product is a fixed number c.
So, we have a1,a2,a3,....,an=c{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}=c
Now, we will multiply both sides by 2 in the above equation, that means, we get,
⇒a1,a2,a3,....,2an=2c\Rightarrow {{a}_{1}},{{a}_{2}},{{a}_{3}},....,2{{a}_{n}}=2c… (i)
As we know that for n positive numbers the arithmetic mean is greater than or equal to geometric mean, that means, we get,
Arithmetic Mean ≥\ge Geometric Mean
⇒a1+a2+a3+.....+an−1+2ann≥a1,a2,a3,....(2an)n\Rightarrow \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.....+{{a}_{n-1}}+2{{a}_{n}}}{n}\ge \sqrt[n]{{{a}_{1}},{{a}_{2}},{{a}_{3}},....(2{{a}_{n}})}
From the equation (i), we can get,
⇒a1+a2+a3+.....+an−1+2ann≥2cn\Rightarrow \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.....+{{a}_{n-1}}+2{{a}_{n}}}{n}\ge \sqrt[n]{2c}
Now, we will take the n from left-hand side to right-hand side and it will be in the form of multiplication on the right-hand side. Also, we will rearrange n square root to1n\dfrac{1}{n}, that means, we get,
⇒a1+a2+a3+.....+an−1+2an≥n.(2c)1n\Rightarrow {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.....+{{a}_{n-1}}+2{{a}_{n}}\ge n.{{(2c)}^{\dfrac{1}{n}}}
Therefore, we get the minimum value ofa1+a2+a3+......an−1+2an{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......{{a}_{n-1}}+2{{a}_{n}} is n.(2c)1nn.{{(2c)}^{\dfrac{1}{n}}}.

So, the correct answer is option A.

Note:
So, from the given problem statement we learnt the concept of the arithmetic mean and geometric mean. We just need to note that whenever there are n positive numbers at that point of time the arithmetic mean is greater than or equal to the geometric mean.