How do I prove this using question using Cauchy-Schwarz inequality? $$(a_1 + a_2 + · · · + a_n)( 1/a_1 +1/a_2+ · · · +1/a_n) ≥ n^2$$
My TA mentioned he solved this using Cauchy Schwartz inequality, but I don't see how. All I've managed to do is make a summation of $1$, $n$ times, which ends up being $n ≥ n^2$.
 A: Assuming $a_j>0$, it's the straight-forward application of Cauchy-Schwartz to the two vectors $$(\sqrt{a_1},\cdots,\sqrt{a_n});\qquad \left(\frac1{\sqrt{a_1}},\cdots,\frac1{\sqrt{a_n}}\right)$$
A: Let $a_1$, $a_2$, $\ldots,\,a_n$ real numbers greater than $0$.
Since
\begin{align*}\left(a_1+a_2+\ldots+a_n\right)\left(\frac1{a_1}+\frac1{a_2}+\ldots+\frac1{a_n}\right)&=n+\sum_{i=1\\j>i}^n\left(\frac{a_i}{a_j}+\frac{a_j}{a_i}\right)
\end{align*}
(Observe that the last sum contains ${ n\choose 2}$ terms of the form $\frac{a_i}{a_j}+\frac{a_j}{a_i}$). It is only needed to prove that $$\sum_{i=1\\j>i}^n\left(\frac{a_i}{a_j}+\frac{a_j}{a_i}\right)\ge n(n-1)$$
which follows from the AM-GM inequality: $$\frac{a_i}{a_j}+\frac{a_j}{a_i}\ge 2\sqrt{\frac{a_i}{a_j}\cdot\frac{a_j}{a_i}}=2\quad\text{for all }i\text{ and }j.$$
A: Since you haven't accepted the other answers I'll assume that you want more detail. First, a quick statement about the Cauchy-Schwarz inequality. Stated directly, for two vectors $u$ and $v$ of an inner product space $$|\langle u,v\rangle|^2 \leq \langle u,u \rangle\cdot  \langle v,v\rangle .$$
Now, if we have vectors in $\mathbb{R}^n$ and use the dot product we have an inner product space. In particular, if $u = (u_1,u_2,\ldots,u_n), v = (v_1,v_2,\ldots,v_n),$ then the Cauchy-Schwarz inequality tells us that $$u_1v_1+u_2v_2+\ldots+u_nv_n \leq \left(\sqrt{u_1^2+u_2^2+\ldots+u_n^2}\right)\left(\sqrt{v_1^2+v_2^2+\ldots+v_n^2}\right).$$
Now, in this particular case, consider $$u = \left(\sqrt{a_1},\sqrt{a_2},\ldots,\sqrt{a_n}\right), \quad v = \left(\frac{1}{\sqrt{a_1}},\frac{1}{\sqrt{a_2}},\ldots,\frac{1}{\sqrt{a_n}}\right).$$
Applying the above form of the Cauchy-Schwarz inequality with the dot product, we get $$\frac{\sqrt{a_1}}{\sqrt{a_1}},\frac{\sqrt{a_2}}{\sqrt{a_2}},\ldots,\frac{\sqrt{a_2}}{\sqrt{a_n}} = n \leq \left(\sqrt{a_1+a_2+\ldots+a_n}\right)\left(\sqrt{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}}\right).$$ Next, we simply square both sides to arrive at $$n^2 \leq \left(a_1+a_2+\ldots+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\right),$$ as desired. Let me know if any of this wasn't clear.
A: Hint (assuming $a_k \gt 0\,$): by the harmonic mean inequality:
$$
\frac{\sum_{k=1}^n a_k}{n} \;\ge\; \frac{n}{\;\;\sum_{k=1}^n\cfrac{1}{a_k}\;\;}
$$
