Question

If \[a,\,\,b,\,\,c>0\] and \[abc=1\], then the value of \[a+b+c+ab+bc+ca\] lies in the interval
A) \[(\infty ,-6)\]
B) \[(-6,0)\]
C) \[(0,6)\]
D) \[[6,\infty )\]

Answer 1

Hint: In this particular problems there are two conditions are given \[a,\,\,b,\,\,c>0\] and \[abc=1\] we have to find the values of \[a+b+c+ab+bc+ca\] that means we have to use the conditions that is \[A.M\ge G.M\] and \[G.M\ge H.M\] where, \[A.M\] stands for Arithmetic Mean, \[G.M\] stands for Geometric Mean, and \[H.M\] stands for Harmonic Mean.

Complete step by step answer:
Here, in this problems given conditions are \[a,\,\,b,\,\,c>0\] and \[abc=1\]
And in question it is asked to find the value of \[a+b+c+ab+bc+ca\]
To find the values of above equation
First of all we need to use the condition in which Arthematic mean greater than or equal to Geometric Mean
\[A.M\ge G.M\] Where, \[A.M\] stands for Arithmetic Mean, \[G.M\] stands for Geometric Mean
As, we know Arthematic mean is given by \[A.M=\dfrac{a+b+c}{3}\]
And geometric mean is given by \[G.M=\sqrt[3]{abc}\]
By substituting the value of \[A.M\] and \[G.M\] on this condition \[A.M\ge G.M\]we get:
\[\dfrac{a+b+c}{3}\ge \sqrt[3]{abc}\]
As, we know that condition which is given in the question is that \[abc=1\]
Therefore, we get:
\[\dfrac{a+b+c}{3}\ge \sqrt[3]{1}\]
By simplifying we get:
\[a+b+c\ge 3--(1)\]

Another condition is that
\[G.M\ge H.M\] Where, \[G.M\] stands for Geometric Mean, and \[H.M\] stands for Harmonic Mean.
As, we know geometric mean is given by \[G.M=\sqrt[3]{abc}\]
And Harmonic mean is given by \[H.M=\dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\]
Substitute the value of \[G.M\]and \[H.M\]on this condition \[G.M\ge H.M\]we get:
\[\sqrt[3]{abc}\ge \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\]
As, we know that condition which is given in the question is that \[abc=1\]
Substitute in above equation we get:
\[\sqrt[3]{1}\ge \dfrac{3}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\]
By simplifying and taking LHS we get:
\[1\ge \dfrac{3}{\dfrac{bc+ac+ab}{abc}}\]
By further simplifying we get:
\[1\ge \dfrac{3abc}{bc+ac+ab}\]
As, we know that condition which is given in the question is that \[abc=1\]
\[1\ge \dfrac{3}{bc+ac+ab}\]
By further simplifying we get:
\[bc+ac+ab\ge 3\]
By rearranging the term we get:
\[ab+bc+ac\ge 3--(2)\]
By adding the equation (1) and equation (2) and further simplifying we get:
\[a+b+c+ab+bc+ac\ge 6\]
If you observe carefully then you can notice that the value which we get is greater than or equal to 6. That means the value of \[a+b+c+ab+bc+ca\] lies in the interval of \[[6,\infty )\].
So, the correct option is “option (D)”.

Note:
In this particular case you have to always remember two conditions that mean \[A.M\ge G.M\] as well as \[G.M\ge H.M\]. Don’t make silly mistakes while simplifying and also substituting the values in the \[a+b+c+ab+bc+ca\]. If you notice the Harmonic means then it is inverse proportional to the Arithmetic mean. The value which we get is greater than or equal to 6 the meaning of this statement value will be more or equal to 6 and reach to\[\infty \]. So, the above solution is referred to for such types of problems.