Given $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$ and $b_1$, $b_2$, $b_3$, $\ldots$, $b_n$ positive real numbers such that $$a_1 + a_2 + a_3 +\cdots+ a_n \geq a_1b_1+a_2b_2+a_3b_3...a_nb_n$$ show that: $$a_1 + a_2 + a_3+\cdots+a_n \leq \frac{a_1}{b_1} + \frac{a_2}{b_2} + \frac{a_3}{b_3}+\cdots+\frac{a_n}{b_n}$$

This was an admissions question for a math camp I ended up going to. After the program, I'm still wondering how to prove this. (The hint I got at the camp was to use Cauchy-Schwarz, but not sure where.)

I would love hearing solutions both with and without C-S, hope you can help!

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    $\begingroup$ The given inequality can be rewritten as $a_1(1-b_1)+a_2(1-b_2)+...\geq0$. The second can be rewritten as $a_1(\frac1{b_1}-1)+a_2(\frac1{b_2}-1)+...\geq0$. $\frac1{b_1}-1=\frac{1-b_1}{b_1}$, so we seek to show that $a_1\cdot\frac{1-b_1}{b_1}+a_2\cdot\frac{1-b_2}{b_2}+...\geq0$. From here, you can apply Cauchy-Schwarz... $\endgroup$ Commented Aug 14, 2019 at 20:58
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    $\begingroup$ Sorry, I'm still not clear where Cauchy-Schwarz comes into play $\endgroup$ Commented Aug 14, 2019 at 21:24
  • $\begingroup$ Think about this: A given term in the desired inequality is negative iff $b_i>1$. If $b_i>1$, then $$|a_i\cdot\frac{1-b_i}{b_i}|<|a_i(1-b_i)|$$but clearly both are negative, so $$a_i\cdot\frac{1-b_i}{b_i}>a_i(1-b_i)$$If $b_i<1$, then clearly $$a_i\cdot\frac{1-b_i}{b_i}>a_i(1-b_i)$$ This means that if $b_i\neq1$, $$a_i\cdot\frac{1-b_i}{b_i}>a_i(1-b_i)$$If $b_i=1$, $$a_i\cdot\frac{1-b_i}{b_i}=a_i(1-b_i)=0$$ $\endgroup$ Commented Aug 14, 2019 at 22:04
  • $\begingroup$ If we know that $$\sum_{k=1}^n a_k \geq \sum_{k=1}^n a_k b_k$$ why can't we immediately conclude that $$\sum_{k=1}^n \frac{a_k}{b_k} \geq \sum_{k=1}^n a_k$$ provided that $b_k \neq 0$? I had originally typed up an answer but realized that my reasoning was probably wrong as I didn't apply the CS inequality. $\endgroup$
    – Axion004
    Commented Aug 15, 2019 at 0:04

3 Answers 3


By C-S $$\sum_{k=1}^n\frac{a_k}{b_k}\sum_{k=1}^na_kb_k\geq\left(\sum_{k=1}^na_k\right)^2\geq\sum_{k=1}^na_k\sum_{k=1}^na_kb_k$$ and we are done!

  • $\begingroup$ In order to form $$\sum_{k=1}^n\frac{a_k}{b_k}\sum_{k=1}^na_kb_k\geq\left(\sum_{k=1}^na_k\right)^2$$ Did you originally start at $$\sum_{k=1}^n\Big(\frac{a_k}{b_k}\Big)^2\sum_{k=1}^n\Big({a_k}{b_k}\Big)^2\geq\left(\sum_{k=1}^n (a_k)^2\right)^2$$ and take square roots? I'm curious due to the general form of C-S $$\bigg(\sum_{k=1}^n{a_k}^2\bigg)\bigg(\sum_{k=1}^n {b_k}^2\bigg)\geq\left(\sum_{k=1}^n a_k b_k\right)^2$$ $\endgroup$
    – Axion004
    Commented Aug 15, 2019 at 2:54
  • $\begingroup$ @Axion004 I think it would be another problem. $\endgroup$ Commented Aug 15, 2019 at 3:40

I think this will work. I'm concerned that I didn't apply the CS inequality and therefore made a mistake near the last step.

As discussed in the comments

$$a_1 + a_2 + a_3...a_n \geq {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}...{a_n}{b_n}$$

can be rewritten as

$$a_1\Big(\frac{1-b_1}{b_1}\Big)+a_2\Big(\frac{1-b_2}{b_2}\Big)+\dots+a_n\Big(\frac{1-b_n}{b_n}\Big)\geq 0$$

then as $a_i,b_i (i=1,2,\dots n)$ are positive real numbers, we need to consider what happens to


by analyzing three different cases.

Case 1: $~0 < a_i,b_i < 1$. In this case,

$$a_i\Big(\frac{1-b_i}{b_i}\Big) >a_i\Big({1-b_i}\Big)$$

since $0 < b_i < 1$.

Case 2: $a_i=b_i=1$. In this case,

$$a_i\Big(\frac{1-b_i}{b_i}\Big) =a_i\Big({1-b_i}\Big)=0$$

because $b_i = 1$.

Case 3: $~a_i,b_i > 1$. In this case,

$$a_i\Big(\frac{1-b_i}{b_i}\Big) >a_i\Big({1-b_i}\Big)$$

because both are negative but the LHS will be a smaller negative number.

This allows us to conclude that

$$\sum_{i=1}^n a_i\Big(\frac{1-b_i}{b_i}\Big) > \sum_{i=1}^n a_i\Big({1-b_i}\Big)$$

which can be rearranged to (provided $b_i \neq 1$ in which case the sum is $0$)

$$\sum_{i=1}^n \bigg(\frac{a_i\Big(\frac{1-b_i}{b_i}\Big)}{1-b_i}\bigg) > \sum_{i=1}^n a_i$$

so that

$$\sum_{i=1}^n a_i < \sum_{i=1}^n \Big(\frac{a_i}{b_i}\Big)$$

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    $\begingroup$ What happens if a_i>1, a_j<1,b_k>1,b_l<1? $\endgroup$ Commented Aug 15, 2019 at 13:37

(I consider this different from Michael's solution, but let me know if that's not the cse.)

Proof by contradiction.

Suppose not, so $ \sum a_i > \sum \frac{a_i}{b_i}$.
Then we have the following contradiction:

$$ \sum a_i b_i \sum \frac{a_i}{b_i} \geq (\sum a_i)^2 > \sum a_i b_i \sum \frac{a_i }{b_i }$$


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