Question

Show that $1\le {{\left( 1+{{x}^{3}} \right)}^{\dfrac{1}{2}}}\le 1+{{x}^{3}}$ for $x\ge 0$ ?

Answer 1

Hint: In this question, we have to prove an inequation. In inequation, the terms are not equal to each other but are either greater than or less than equal to each other. So, we solve this problem by using $x\ge 0$ . Then, we will add 1 on both sides of the inequation and thus prove our 1st inequation Then, we will let $1+{{x}^{3}}=u$ and take the square root, thus proving the 2nd inequation. After that, we will prove that the square root of any number is always less than that number, which gives the 3rd inequation to be proved. Hence, in the end, we combine them all to get the required result for the problem.

Complete step by step answer:
According to the question, we have to prove an inequation.
The inequation given to us is $1\le {{\left( 1+{{x}^{3}} \right)}^{\dfrac{1}{2}}}\le 1+{{x}^{3}}$ for $x\ge 0$ .
As we know $x\ge 0$ --------- (1)
Now, we will take cube on both sides of the inequation (1), we get
${{x}^{3}}\ge {{0}^{3}}$
Now, add 1 on both sides in the above inequation, we get
$1+{{x}^{3}}\ge 1+0$
Therefore, we get
$1\le 1+{{x}^{3}}$ ------- (2)
Now, we will let $1+{{x}^{3}}=u$ , therefore take the square root on both sides of the inequation, we get
$\sqrt{\left( 1+{{x}^{3}} \right)}=\sqrt{u}$
As we know, the square root is an increasing function, which implies the square root of a positive number is always greater than 1, that is
$\sqrt{\left( 1+{{x}^{3}} \right)}=\sqrt{u}\ge 1$
Therefore, we get
$1\le \sqrt{\left( 1+{{x}^{3}} \right)}$ --------- (3)
As we know, the square root of a number is always less than or equal to that number, which means the square root of u is always less than or equal to u, that is
$\sqrt{u}\le u$
Therefore, we get
$\sqrt{\left( 1+{{x}^{3}} \right)}\le 1+{{x}^{3}}$ ------ (4)
Thus from equations (2), (3), and (4), we get that $1\le {{\left( 1+{{x}^{3}} \right)}^{\dfrac{1}{2}}}\le 1+{{x}^{3}}$ for $x\ge 0$ , which is our required answer to the problem.

Note:
While solving this problem, keep in mind the technique you are using for solving this question, that is always start your problem using $x\ge 0$ . Also, remember that in the given inequation, we have to prove 3 inequalities, that $\sqrt{\left( 1+{{x}^{3}} \right)}\le 1+{{x}^{3}}$ , $1\le \sqrt{\left( 1+{{x}^{3}} \right)}$, and $1\le 1+{{x}^{3}}$ , to get the accurate answer.