An Approach through Inequalities with SMO 2013 - Round 2 (Senior) - P3

Let $b_1, b_2,\ldots,b_n$ be a sequence of positive real numbers such that for each $n\geq 1$ ,

$b_{n+1}^2 \geq \frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+\cdots+\frac{b_n^2}{n^3}$

Show that there is a positive integer $M$ such that

$\sum_{n=1}^M \frac{b_{n+1}}{b_1+b_2+\cdots+b_n}>\frac{2013}{1013}$

Think about this problem yourself before scrolling down to the solution.......

At the first glance, an obvious idea would be dividing both sides of given inequality by $(b_1+b_2+\cdots+b_n)^2$ to get,

$\left(\frac{b_{n+1}}{b_1+b_2+\cdots+b_n}\right)^2 \geq \left(\frac{b_1}{b_1+b_2+\cdots+b_n}\right)^2\frac{1}{1^3}+\left(\frac{b_2}{b_1+b_2+\cdots+b_n}\right)^2\frac{1}{2^3}+\cdots$

$+\left(\frac{b_n}{b_1+b_2+\cdots+b_n}\right)^2\frac{1}{n^3}$

$\Rightarrow \left(\frac{b_{n+1}}{b_1+\cdots+b_n}\right) \geq \sqrt{\left(\frac{b_1}{b_1+\cdots+b_n}\right)^2\frac{1}{1^3}+\cdots+\left(\frac{b_n}{b_1+\cdots+b_n}\right)^2\frac{1}{n^3}}$

But now we are stuck in reducing the RHS further and we are unable to analyse it properly.

So let us attack this problem from a different angle. There is a magical identity that separates $1^3,2^3,\ldots,n^3$ from $b_1^2,b_2^2,\ldots,b_n^2$ in the given inequality by summing them which is known as the Cauchy Schwarz Inequality. It states that if $x_1,x_2,\ldots,x_n$ and $y_1,y_2,\ldots,y_n$ are two real sequences then

$(x_1^2+x_2^2+\cdots+x_n^2)(y_1^2+y_2^2+\cdots+y_n^2)\geq (x_1y_1+x_2y_2+\cdots+x_ny_n)^2$

where the equality holds if and only if $\frac{x_1}{y_1}=\frac{x_2}{y_2}=\cdots=\frac{x_n}{y_n}$ for no $y_i's$ being zero and if some $y_i=0$ , then $y_i=0\ \forall i\in\{1,2,\ldots,n\}$ . There are several proofs of the Cauchy Schwarz Inequality. Try to prove it in your own way and write the same in the comments !

Now, let the two real sequences be $\left(\frac{b_1}{1^{\frac{3}{2}}}, \frac{b_2}{2^{\frac{3}{2}}},\ldots,\frac{b_n}{n^\frac{3}{2}}\right)$ and $\left(1^{\frac{3}{2}},2^{\frac{3}{2}},\ldots,n^{\frac{3}{2}}\right)$ . Hence by the Cauchy Schwarz we get,

$\left(\frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+\cdots+\frac{b_n^2}{n^3}\right)\left(1^3+2^3+\cdots+n^3\right)\geq (b_1+b_2+\cdots+b_n)^2$

$\Rightarrow \left(\frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+\cdots+\frac{b_n^2}{n^3}\right)\left(\frac{n(n+1)}{2}\right)^2\geq (b_1+b_2+\cdots+b_n)^2$

$\Rightarrow b_{n+1}^2\left(\frac{n(n+1)}{2}\right)^2\geq \left(\frac{b_1^2}{1^3}+\frac{b_2^2}{2^3}+\cdots+\frac{b_n^2}{n^3}\right)\left(\frac{n(n+1)}{2}\right)^2 \geq (b_1+b_2+\cdots+b_n)^2$

$\Rightarrow \frac{b_{n+1}}{b_1+b_2+\cdots+b_n}\geq \frac{2}{n(n+1)}\quad (\text{Rearranging the terms, note $b_i's>0$})$

$\Rightarrow \frac{b_{n+1}}{b_1+b_2+\cdots+b_n}\geq \frac{2}{n}-\frac{2}{n+1}$

$\Rightarrow \boxed{\sum_{n=1}^{N} \frac{b_{n+1}}{b_1+b_2+\cdots+b_n}\geq \frac{2}{1}-\frac{2}{N+1}}\quad (\text{By Telescoping Sum})$

Note that $\frac{2013}{1013}=2-\frac{13}{1013}$ . Hence, if we take $N=1013$ , then $\sum_{n=1}^{N} \frac{b_{n+1}}{b_1+b_2+\cdots+b_n}\geq 2-\frac{2}{N+1}>2-\frac{13}{1013}=\frac{2013}{1013}$ . Hence proved.

Does the equality hold true for the boxed inequality?

If yes, write the required conditions on $b_1,\ldots,b_n$ and if no, then write the reason for it in the comments !

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An Approach through Inequalities with SMO 2013 - Round 2 (Senior) - P3

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