Math, asked by duragpalsingh, 1 year ago

\textsf{Let $a_1, a_2,....,a_n \  \textgreater \ 0,\dfrac{1}{a_1}+\dfrac{1}{a_2}+.....+\dfrac{1}{a_n} = 1, n \in \mathbb{N} - \{0\}$}. \\\\\sf Prove\  that:\\\\ \displaystyle n + \sqrt[3]{\dfrac{a^3_1+a^3_2}{2}} + \sqrt[3]{\dfrac{a^3_2+a^3_3}{2}} + \sqrt[3]{\dfrac{a^3_3+a^3_1}{2}} \leq 2(a_1 + a_2+.......+a_n)

Answers

Answered by devanayan2005
1

Lets simplify our problem and take n=2.

let a1 = 2 , then a2 = 1/2

or in general if a1 = k, then a2 = 1/k, such that a1.a2 = 1

Now if we take the function f = (1+a1+(a1^2))(1+a2+(a2^2))

Put a1 = k, a2 = 1/k , we get

f = ((1+k+(k^2))/k)^2

Now, we can extend this model to n.

All we have to do is, understand that a1*a2*a3*…*an = 1,

whenever numerator is equal to denominator.

for example, consider the sequence

2,3,4,1/2,1/3,1/4, this if taken the product leaves 1.

2*3*4 = 24 and 1/2*1/3*1/4 = 1/*24.

we can assume the numerator equal denominator to be K

thus, f = (((1+K+(K^2))/k)^n

here, we see its an increasing function and given the domain of positive real numbers this function has a minimum at k = 1. The minimum value of the function will be 3^n.

Minimum of f = 3^n.

I Hope it helps, I am learning to type in Latex. :)

Answered by Anonymous
2

Step-by-step explanation:

Refer to the attachmmet

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