Prove $\sum_{k=1}^n\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^n\frac{b_i^2}{a_i}$

Question

How do I prove this? $$\sum_{k=1}^n\frac{(b_1+b_2+\cdots+b_k)b_k}{a_1+a_2+\cdots+a_k}<2\sum_{i=1}^n\frac{b_i^2}{a_i}$$ Here $a_i,b_i\in\Bbb R^+$. I guess the sum transform works, but I can't prove it.

Answer 1

Let $B_k=b_1+\ldots +b_k$ and $A_k=a_1+\ldots+a_k$. Moreover, let $a_0=b_0=A_0=B_0=0$.

We may notice that: $$ B_n^2-B_{n-1}^2 = (B_n-B_{n-1})(B_n+B_{n-1}) = b_n (2B_n-b_n) = 2b_n B_n - b_n^2 \tag{1}$$ hence: $$ \sum_{k=1}^{n}\frac{B_k b_k}{A_k} = \frac{1}{2}\sum_{k=1}^{n}\frac{B_k^2-B_{k-1}^2}{A_k}+\frac{1}{2}\sum_{k=1}^{n}\frac{b_k^2}{A_k}.\tag{2} $$ By Titu's lemma we have: $$ \sum_{k=1}^{m}\frac{b_k^2}{a_k}\geq \frac{B_m^2}{A_m} \tag{3}$$ and by summation by parts we have: $$\sum_{k=1}^{n}\frac{B_k^2-B_{k-1}^2}{A_k} = \frac{B_n^2}{A_n}+\sum_{k=1}^{n-1}\frac{B_k^2 a_{k+1}}{A_k A_{k+1}}\tag{4} $$ The claim can be proved by $(2),(3),(4)$ and induction on $n$.

Answer 2

Here is An answer by toshihiro shimizu on AoPS:

I considered this problem long time and finally I found the following inductive solution. I'd like to know the direct solution.

We show the following stronger result by induction; $$\begin{align*} \sum_{k=1}^{n} & \frac{\left(b_{1}+b_{2}+\cdots+b_{k}\right)b_{k}}{a_{1}+a_{2}+\cdots+a_{k}}\leq\frac{3}{2}\frac{b_{1}^{2}}{a_{1}}+2\sum_{k=2}^{n}\frac{b_{k}^{2}}{a_{k}}. \end{align*}$$ $n=1$ is obvious. We assume the result of $n=1$ and we show the result of $n$. Applying the inductive result to $\left(a_{1}+a_{2},a_{3},a_{4}\ldots,a_{n},b_{1}+b_{2},b_{3},b_{4},\ldots,b_{n}\right)$

we have $$\begin{align*} \frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}} & +\sum_{k=3}^{n}\frac{\left(b_{1}+\cdots+b_{k}\right)b_{k}}{a_{1}+\cdots+a_{k}}\leq\frac{3}{2}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}}+2\sum_{k=3}^{n}\frac{b_{k}^{2}}{a_{k}} \end{align*}$$ It's sufficient to show that $$\begin{align*} \frac{b_{1}^{2}}{a_{1}}+\frac{\left(b_{1}+b_{2}\right)b_{2}}{a_{1}+a_{2}}+\frac{1}{2}\frac{\left(b_{1}+b_{2}\right)^{2}}{a_{1}+a_{2}} & \leq\frac{3}{2}\frac{b_{1}^{2}}{a_{1}}+2\frac{b_{2}^{2}}{a_{2}}\\ \frac{b_{1}^{2}+4b_{1}b_{2}+3b_{2}^{2}}{a_{1}+a_{2}} & \leq\frac{b_{1}^{2}}{a_{1}}+4\frac{b_{2}^{2}}{a_{2}}\\ b_{1}^{2}+4b_{1}b_{2}+3b_{2}^{2} & \leq b_{1}^{2}+\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2}+4b_{2}^{2}\\ 4b_{1}b_{2} & \leq\frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2}+b_{2}^{2} \end{align*}$$ The last inequality is true since $$\begin{align*} \frac{a_{2}}{a_{1}}b_{1}^{2}+4\frac{a_{1}}{a_{2}}b_{2}^{2} & \geq2\sqrt{\frac{a_{2}}{a_{1}}b_{1}^{2}\cdot4\frac{a_{1}}{a_{2}}b_{2}^{2}}\\ & =4b_{1}b_{2} \end{align*}$$ , completing the proof.