# Inequality from Analysis Qual [closed]

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?

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

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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.

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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 $x_iy_i=1$.

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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.

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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.

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