# A better way to prove this inequality [duplicate]

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Exercise 1.1.6. (b) For positive real numbers $a_1, a_2, ... , a_n$ prove that

$$(a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) \geq n^2.$$

From $AM \geq GM$:

$$a_1+a_2+\ldots+a_n \geq n\sqrt[n]{a_1a_2 \ldots a_n}$$ $$a_2a_3 \ldots a_n + a_1a_3 \ldots a_n + \ldots + a_1a_2 \ldots a_{n-1} \geq n\sqrt[n]{(a_1a_2 \ldots a_n)^{n-1}}$$

Multiply, $$(a_1+a_2+\ldots+a_n)(a_2a_3 \ldots a_n + a_1a_3 \ldots a_n + \ldots + a_1a_2 \ldots a_{n-1}) \geq n^2 \cdot a_1a_2 \ldots a_n$$

$$(a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) \geq n^2.$$

I'm not satisfied with this. I'm looking for a better way to prove it...or at least a better way to write it (notation).

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• Cauchy-Schwarz? – Lord Shark the Unknown Mar 17 '18 at 7:15
• – Martin R Mar 17 '18 at 7:47

That's just the AM-HM inequality written differently:

$$\frac{a_1+a_2+ \ldots +a_n}{n} \ge \frac{n}{\cfrac{1}{a_1}+\cfrac{1}{a_2}+ \ldots +\cfrac{1}{a_n}}$$

Since our inequality is homogeneous, we can assume that $$a_1+a_2+...+a_n=n$$ and we need to prove that $$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\geq n,$$ which is true because $$\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}-n=\sum_{i=1}^n\left(\frac{1}{a_i}-1\right)=$$ $$=\sum_{i=1}^n\left(\frac{1}{a_i}-1+a_i-1\right)=\sum_{i=1}^n\frac{(a_i-1)^2}{a_i}\geq0.$$

\begin{align*} n^{2}&=\left(\sqrt{a_{1}}\cdot\dfrac{1}{\sqrt{a_{1}}}+\cdots+\sqrt{a_{n}}\cdot\dfrac{1}{\sqrt{a_{n}}}\right)^{2}\\ &\leq\left(a_{1}+\cdots+a_{n}\right)\left(\dfrac{1}{a_{1}}+\cdots+\dfrac{1}{a_{n}}\right) \end{align*} by Cauchy-Schwarz.

Cauchy-Schwarz:

$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\ge\left[\sqrt{a_1}\cdot\sqrt{\frac{1}{a_1}}+\sqrt{a_2}\cdot\sqrt{\frac{1}{a_2}}+\cdots+\sqrt{a_n}\cdot\sqrt{\frac{1}{a_n}}\right]^2$$

A.M.-G.M.

$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\ge\left(n\cdot\sqrt[n]{a_1a_2\cdots a_n}\right)\left(n\cdot\sqrt[n]{\frac{1}{a_1a_2\cdots a_n}}\right)$$

Cauchy-Schwarz: $$(a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) = ((\sqrt{a_1})^2+(\sqrt{a_2})^2+ \ldots +(\sqrt{a_n})^2)\Big(\frac{1}{(\sqrt{a_1})^2}+\frac{1}{(\sqrt{a_2})^2}+ \ldots +\frac{1}{(\sqrt{a_n})^2}\Big) \geq (1 + 1 + \ldots + 1)^2 = n^2.$$

Nothing except $a+\frac 1a\geqslant 2\sqrt{a\cdot\frac 1a}=2$ for $a,b>0:$

$\sum\limits_{k=1}^n a_k\cdot \sum\limits_{m=1}^n \frac1{a_m}=\sum\limits_{k=m=1}^n 1+\sum\limits_{1\leqslant k<m\leqslant n} \left(\frac{a_m}{a_k}+\frac{a_k}{a_m}\right)\geqslant n + \sum\limits_{1\leqslant k<m\leqslant n} 2=n+\binom {n}{2}\cdot 2=n^2$