# Problem on Inequality | ISI - MSQMS - B, 2018 | Problem 4b

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.

## INEQUALITY | ISI 2018| MSQMS | PART B | PROBLEM 4b

Let $a>0$ and $n \in \mathbb{N} .$ Show that
"
$$\frac{a^{n}}{1+a+a^{2}+\ldots+a^{2 n}}<\frac{1}{2 n}$$

Algebra

Inequality

Numbers

## Try with Hints

We know if $\frac{a}{b} \geq c$

Then $\frac{b}{a} \leq c$

$\frac{a^n}{1+a+a^2+.......+a^{2n}}$ < $\frac{1}{2n}$

$\frac{1+a+a^2+.....+a^{2n}}{a^n}$ > $2n$

Now why don't you just give it a try yourself,try to conclude something from the previous line,I know you can.

Therefore, $1+(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+.............+(a^n+\frac{1}{a^n})> 2n$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

$a+\frac{1}{a} \geq 2$

$a^2+\frac{1}{a^2} \geq 2$

.

.

.

$a^n+\frac{1}{a^n} \geq 2$

Adding the above inequalities we get,$(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+..........+(a^n+\frac{1}{a^n}) \geq 2n$

We are almost there ,so just try the last step yourself.

Therefore, $1+(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+..........+(a^n+\frac{1}{a^n}) \geq 2n+1$ > $2n$

Thus, $\frac{a^n}{1+a+a^2+......+a^n}$<$\frac{1}{2n}$

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