# Q: Inequality $\frac{\sum_{k=1}^{M} A_{k}}{\sum_{k=1}^{M} B_{k}} \leq \frac{1}{M} \sum_{k=1}^{M} \frac{A_{k}}{B_{k}}$, where $A_{k}, B_{k} \geq 0$

I wanted to use the following inequality in my research, but I cannot prove whether it is correct or not.

$$\frac{\sum_{k=1}^{M} A_{k}}{\sum_{k=1}^{M} B_{k}} \leq \frac{1}{M} \sum_{k=1}^{M} \frac{A_{k}}{B_{k}}$$, where $$A_{k}, B_{k} \geq 0$$

I tested this inequality on random numbers generated by MATLAB and the inequality seemed to hold. Anyone has some ideas, how to prove or disprove it? Thank you guys in advance.

• Are there any restrictions on $A_k$ and $B_k$ ? – learner Mar 3 '19 at 4:50

The edited version is false: take $$M=2$$ and let $$A_1 \to 0$$. The inequality becomes $$\frac {A_2} {B_1+B_2} \leq \frac 1 2\frac {A_2} {B_2}$$. But this is false if $$B_1 .

Answer for the old version: assuming that $$A_k$$'s and $$B_k$$'s are positive there is a stronger inequality: let $$C$$ be the maximum of the numbers $$\frac {A_k} {B_k}$$. Then $$\sum A_k \leq C\sum b_k$$ so $$\frac {\sum A_k} {\sum B_k} \leq C \leq \sum \frac {A_k} {B_k}$$.

• Sorry for missing the multiplier. I meant $\frac{1}{M}$ instead of M. Just updated the question. – Scott Guan Mar 3 '19 at 5:21
• @ScottGuan I have updated the answer. – Kavi Rama Murthy Mar 3 '19 at 5:37
• Thank you for your answer. Appreciate it. – Scott Guan Mar 3 '19 at 15:06

It's false generally, but true in some cases. I don't have a full characterization of when it's true. Consider Titu's Lemma (a consequence of the Cauchy-Schwartz inequality. See the Cauchy-Schwartz wikipedia page).

$$\frac{(\sum_{k=1}^M u_k)^2}{\sum_{k=1}^M v_k} \leq \sum_{k=1}^M\frac{u_k^2}{v_k}.$$

If $$A_k = u_k = 1$$ and $$B_k = v_k$$ for all $$k$$, then we're done. This implies

$$M\frac{M}{\sum_{k=1}^M B_k} \leq \sum_{k=1}^M \frac{1}{B_k}.$$

Now for a specific counterexample. Let $$A_1 = 20200$$ and $$A_k=1$$ for $$k=2,\dots,11$$. Let $$B_1 = 10000$$ and $$B_k = 10$$ for $$k=2,\dots,11$$. Of course, $$M=11$$. Then,

\begin{align*} &\frac{\sum_{k=1}^M A_k}{\sum_{k=1}^M B_k} = \frac{20210}{10100} \geq 2.\\ &\frac{1}{M}\sum_{k=1}^M\frac{A_k}{B_k} = \frac{1}{11}\left(2.02 + 10*0.1\right) = \frac{3.02}{11} < 1. \end{align*}

I just chose some extreme numbers to make an example. You could probably construct a cleaner counterexample later. Here's how I constructed it. Start with arbitrary positive $$B_k$$ (assume the $$B_k$$ are not all equal to each other.) and let $$A_k = 1$$ for all $$k$$. In this case, we know the inequality holds. Then take the partial derivatives with respect to $$A_k$$ for some $$k$$ such that $$B_k > \overline{B}:=\frac{1}{M}\sum_{k=1}^M B_k$$. Then the derivative on the left side of the inequality is $$\frac{1}{M\overline{B}}$$ and the derivative on the right side of the inequality is $$\frac{1}{MB_k} < \frac{1}{M\overline{B}}$$. So for a counterexample, make $$A_k$$ large if $$B_k$$ is large.

• Sorry for missing the multiplier. I meant $\frac{1}{M}$ instead of M. Just updated the question. – Scott Guan Mar 3 '19 at 5:20
• Updated the answer. – forgottenarrow Mar 3 '19 at 6:03
• Thanks for the answer. It helped a lot. – Scott Guan Mar 3 '19 at 15:06

Just complementing the other answers. When it works, it is known as Chebyshev's inequality. For example $$A_1\geq A_2\geq ... \geq A_M \geq 0$$ and $$\color{red}{B_M\geq B_{M-1}\geq ... \geq B_1>0}$$ then $$\frac{1}{B_1}\geq \frac{1}{B_2}\geq ... \geq \frac{1}{B_M}>0\Rightarrow \color{red}{\frac{A_1}{B_1}\geq \frac{A_2}{B_2}\geq ... \geq \frac{A_M}{B_M}}$$ as a result $$0\leq M\left(\sum\limits_{k=1}^M A_k\right)= \color{blue}{M\left(\sum\limits_{k=1}^M \frac{A_k}{B_k}\cdot B_k\right)\overset{Ch.in.}{\leq} \left(\sum\limits_{k=1}^M \frac{A_k}{B_k}\right)\left(\sum\limits_{k=1}^M B_k\right)}$$ and finally $$\frac{\sum\limits_{k=1}^M A_k}{\sum\limits_{k=1}^M B_k} \leq \frac{1}{M}\left(\sum\limits_{k=1}^M \frac{A_k}{B_k}\right)$$