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# The least natural number $a$ for which $x + a{x^{ - 2}} > 2,{\rm{ }}\forall x \in \left( {0,\infty } \right)$ is(a) 1 (b) 5 (c) 2 (d) None of these

Last updated date: 11th Sep 2024
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Hint:
Here, we need to find the least value of $a$. We will rewrite the number $x$ as the sum of $\dfrac{x}{2}$ and $\dfrac{x}{2}$, and rewrite the given inequation. Then, we will use the relation between arithmetic mean and geometric mean to form another inequation. Finally, we will compare the two inequalities, and simplify them to get the least value of the natural number $a$.

Formula Used:
We will use the following formulas:
1) The arithmetic mean of the $n$ numbers ${a_1},{a_2}, \ldots \ldots ,{a_n}$ is given by the formula $A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + {a_n}}}{n}$.
2) The geometric mean of the $n$ numbers ${a_1},{a_2}, \ldots \ldots ,{a_n}$ is given by the formula $G.M. = \sqrt[n]{{{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}}}$.

Complete step by step solution:
We will use the formula for A.M. and G.M. to solve the problem.
Rewriting the equation, we get
$\Rightarrow x + \dfrac{a}{{{x^2}}} > 2$
Rewriting $x$ as the sum of $\dfrac{x}{2}$ and $\dfrac{x}{2}$, we get
$\Rightarrow \dfrac{x}{2} + \dfrac{x}{2} + \dfrac{a}{{{x^2}}} > 2$
The arithmetic mean of the $n$ numbers ${a_1},{a_2}, \ldots \ldots ,{a_n}$ is given by the formula $A.M. = \dfrac{{{a_1} + {a_2} + \ldots \ldots + {a_{n - 1}} + {a_n}}}{n}$.
The number of terms in the sum $\dfrac{x}{2} + \dfrac{x}{2} + \dfrac{a}{{{x^2}}}$ is 3.
Therefore, we get the arithmetic mean of the numbers $\dfrac{x}{2},\dfrac{x}{2},\dfrac{a}{{{x^2}}}$ as
$\Rightarrow A.M. = \dfrac{{\dfrac{x}{2} + \dfrac{x}{2} + \dfrac{a}{{{x^2}}}}}{3}$
Simplifying the expression, we get
$\Rightarrow A.M. = \dfrac{{x + \dfrac{a}{{{x^2}}}}}{3}$
The number of terms in the sum $\dfrac{x}{2} + \dfrac{x}{2} + \dfrac{a}{{{x^2}}}$ is 3.
Therefore, using the formula $G.M. = \sqrt[n]{{{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}}} = {\left( {{a_1}{a_2} \ldots \ldots {a_{n - 1}}{a_n}} \right)^{1/n}}$, we get the geometric mean of the numbers $\dfrac{x}{2},\dfrac{x}{2},\dfrac{a}{{{x^2}}}$ as
$\Rightarrow G.M. = {\left( {\dfrac{x}{2} \times \dfrac{x}{2} \times \dfrac{a}{{{x^2}}}} \right)^{1/3}}$
Simplifying the expression, we get
$\Rightarrow G.M. = {\left( {\dfrac{a}{4}} \right)^{1/3}}$
Now, we know that the arithmetic mean is always greater than or equal to the geometric mean.
Therefore, we get
$\Rightarrow A.M. \ge G.M.$
Substituting $A.M. = \dfrac{{x + \dfrac{a}{{{x^2}}}}}{3}$ and $G.M. = {\left( {\dfrac{a}{4}} \right)^{1/3}}$ in the inequation, we get
$\Rightarrow \dfrac{{x + \dfrac{a}{{{x^2}}}}}{3} \ge {\left( {\dfrac{a}{4}} \right)^{1/3}}$
Multiplying both sides by 3, we get
$\Rightarrow x + \dfrac{a}{{{x^2}}} \ge 3{\left( {\dfrac{a}{4}} \right)^{1/3}}$
We have the inequation $x + \dfrac{a}{{{x^2}}} \ge 3{\left( {\dfrac{a}{4}} \right)^{1/3}}$.
This means that the least value of $x + \dfrac{a}{{{x^2}}}$ is $3{\left( {\dfrac{a}{4}} \right)^{1/3}}$.
Also, we have the inequation $x + \dfrac{a}{{{x^2}}} > 2$.
This means that the value of $x + \dfrac{a}{{{x^2}}}$ is more than 2.
Therefore, we can conclude that the least value of $x + \dfrac{a}{{{x^2}}}$ is more than 2.
Thus, we get the inequation
$\Rightarrow 3{\left( {\dfrac{a}{4}} \right)^{1/3}} > 2$
We will simplify this inequation to get the least natural number value of $a$.
Dividing both sides by 3, we get
$\Rightarrow {\left( {\dfrac{a}{4}} \right)^{1/3}} > \dfrac{2}{3}$
Taking the cubes of both sides, we get
$\begin{array}{l} \Rightarrow \dfrac{a}{4} > {\left( {\dfrac{2}{3}} \right)^3}\\ \Rightarrow \dfrac{a}{4} > \dfrac{8}{{27}}\end{array}$
Multiplying both sides of the inequation by 4, we get
$\begin{array}{l} \Rightarrow \dfrac{a}{4} \times 4 > \dfrac{8}{{27}} \times 4\\ \Rightarrow a > \dfrac{{32}}{{27}}\end{array}$
Simplifying the expression, we get
$\Rightarrow a > 1.185$
Since $a$ is a natural number, the least value of $a$ for which $a > 1.185$ is true is $a = 2$.
Therefore, we get the least value of the natural number $a$ as 2.

Thus, the correct option is option (c).

Note:
We have multiplied and divided both sides of inequalities in the solution without changing the inequality sign. This is because when both sides of an inequation are multiplied or divided by the same positive number, the inequality sign remains unchanged. The inequality sign changes when a negative number is multiplied or divided on both sides of inequation.