Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\{a_j\}_{j=1}^N$ be a finite set of positive real numbers. Suppose

$$\sum_{j=1}^{N} a_j = A,$$ prove

$$\sum_{j=1}^{N} \frac{1}{a_j} \geq \frac{N^2}{A}.$$

Hints on how to proceed?

share|cite|improve this question

closed as off-topic by Matthew Conroy, heropup, Jonas, probablyme, Silvia Ghinassi Feb 27 at 2:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Matthew Conroy, heropup, Jonas, probablyme, Silvia Ghinassi
If this question can be reworded to fit the rules in the help center, please edit the question.

    
How about applying the Cauchy-Schwarz inequality to the inner product of two $N$-dim vectors $(a_1^{1/2}, \cdots, a_N^{1/2})$ and $(a_1^{-1/2}, \cdots, a_N^{-1/2})$? – Sangchul Lee Feb 26 at 18:06

$$N=\sqrt{a_1}.\frac{1}{\sqrt{a_1}}+...+\sqrt{a_n}.\frac{1}{\sqrt{a_n}}\leq \sqrt{a_1+...+a_n}\sqrt{\frac{1}{a_1}+...+\frac{1}{a_n}}$$ Now square both sides of the inequality.

share|cite|improve this answer

Try the Cauchy-Schwarz inequality. This would be a 3 line proof. $$ \left(\sum_{i=1}^Nx_iy_i\right)^2\le\sum_{i=1}^Nx_i^2·\sum_{i=1}^Ny_i^2 $$ Now chose $x_i,y_i$ so that one recognizes the sums in the task and that $x_iy_i=1$.

share|cite|improve this answer
    
Yes, Cauchy-Schwarz + one more hint: think to $(1/\sqrt{a_1},\cdots,1/\sqrt{a_N}$ and $(\sqrt{a_1},\cdots,\sqrt{a_N}$. – JeanMarie Feb 26 at 18:07

Let $g(x) = 1/x, x > 0.$ Then $g$ is convex on $(0,\infty),$ hence by Jensen,

$$ g(\frac{1}{N}\sum_{j=1}^{N}a_n) \le \frac{1}{N}\sum_{j=1}^{N}g(a_n).$$

The inequality falls right out.

share|cite|improve this answer

For positive $a,b$ we have $a/b+b/a= (\sqrt {a/b}-\sqrt {b/a})^2+2\geq 2 .$

Therefore $\sum_1^na_i \sum_1^n 1/a_i= \sum_1^n a_i(1/a_i)+\sum_{1\leq i<j\leq n}(a_i/a_j+a_j/a_i)=n+\sum_{1\leq i<j\leq n}(a_i/a_j+a_j/a_i)\geq n+\sum_{1\leq i<j\leq n}2=n+\binom {n}{2}2=n^2.$ The Cauchy-Schwarz Inequality.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.