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

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

Key Concepts


Algebra

Inequality

Numbers


Try with Hints


First hint

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.

Second hint

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

Third hint

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

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

.

.

.

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

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.

Final Step

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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