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Question: Let us assume, \[\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1\] for \[x > 0,\,y > 0,\,z > 0\]. Now...

Let us assume, 1x+1y+1z=1\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1 for x>0,y>0,z>0x > 0,\,y > 0,\,z > 0. Now, find out which of the following sets could be possible values of (x1)(y1)(z1)(x - 1)(y - 1)(z - 1)?
(This question has multiple correct options)
A. [10,)\left[ {10,\infty } \right)
B. [11,)\left[ {11,\infty } \right)
C. [7,)\left[ {7,\infty } \right)
D. [6,)\left[ {6,\infty } \right)

Explanation

Solution

In order to solve the question, first, we have to apply the concept of Arithmetic Mean \geqslant Geometric Mean. The geometric mean takes several values, multiplies them together, and sets them to the 1nth\dfrac{1}{{{n^{th}}}} power. The arithmetic mean is often known simply as the mean.

Complete answer:
Given 1x+1y+1z=1      (1)\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1\;\;\; \ldots \ldots \left( 1 \right)
Let us assume 1x=a,1y=b\dfrac{1}{x} = a,\dfrac{1}{y} = b and 1z=c\dfrac{1}{z} = c
According to question, abc=1abc = 1
If we take aa and bb,
A.M. = a+b2\dfrac{{a + b}}{2}, and
G.M. = ab\sqrt {ab}
The AM-GM inequality concept says that the arithmetic mean of a list of non-negative real numbers is greater than equal to the geometric mean of the same list.
Applying the same concept to aa and bb, we get,

a+b2ab a+b2ab  \dfrac{{a + b}}{2} \geqslant \sqrt {ab} \\\ \Rightarrow a + b \geqslant 2\sqrt {ab} \\\

Similarly for b,cb,c and a,ca,c, we get,
b+c2acb + c \geqslant 2\sqrt {ac} and a+c2aca + c \geqslant 2\sqrt {ac}
Now, multiplying these three inequalities, we get,
(a+b)(b+c)(a+c)8abc\left( {a + b} \right)\left( {b + c} \right)\left( {a + c} \right) \geqslant 8abc
Substituting the value of a,b,ca,b,c in the original equation (1)\left( 1 \right), we obtain,
(1x+1y)(1y+1z)(1x+1z)8xyz\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right)\left( {\dfrac{1}{y} + \dfrac{1}{z}} \right)\left( {\dfrac{1}{x} + \dfrac{1}{z}} \right) \geqslant \dfrac{8}{{xyz}}
From equation (1)\left( 1 \right), we can write it as,
(11z)(11x)(11y)8xyz\Rightarrow \left( {1 - \dfrac{1}{z}} \right)\left( {1 - \dfrac{1}{x}} \right)\left( {1 - \dfrac{1}{y}} \right) \geqslant \dfrac{8}{{xyz}}
Therefore, we get the final result as (x1)(y1)(z1)8(x - 1)(y - 1)(z - 1) \geqslant 8
So, the possible values here for (x1)(y1)(z1)=[8,)(x - 1)(y - 1)(z - 1) = \left[ {8,\infty } \right)
Hence, the possible values of the given expression can be greater than or equal to 88 and upto \infty
The correct answers are A. [10,)\left[ {10,\infty } \right)and B. [11,)\left[ {11,\infty } \right)

Note:
The geometric mean calculates the mean or average of a series of product values, which considers the effect of compounding. It is used to determine the investment performance, whereas the arithmetic mean calculates the mean by the sum of the total values divided by the number of values. It is an average, a measure of the center of a set of data.