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

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.