Verfication of deduction made using the Cauchy-Schwarz inequality

Question

Is the following proof correct?

Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.

Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the euclidean product in $\textbf{6.4}$. Now let $a,b,c,d$ be arbitrary positive numbers and let $u = (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|)$ and $v = (\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})$ then by appealing to the Cauchy Schwartz inequality we have\begin{align*} |\langle u,v\rangle| &= |4| = |\langle (|\sqrt{a}|,|\sqrt{b}|,|\sqrt{c}|,|\sqrt{d}|),(\frac{1}{|\sqrt{a}|},\frac{1}{|\sqrt{b}|},\frac{1}{|\sqrt{c}|},\frac{1}{|\sqrt{d}|})\rangle|\\ &\leq \sqrt{|\sqrt{a}|^2+|\sqrt{b}|^2+|\sqrt{c}|^2+|\sqrt{d}|^2}\cdot\sqrt{\frac{1}{|\sqrt{a}|^2}+\frac{1}{|\sqrt{b}|^2}+\frac{1}{|\sqrt{c}|^2}+\frac{1}{|\sqrt{d}|^2}}\\ &= \sqrt{(a+b+c+d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})} = \|u\|\cdot\|v\|. \end{align*} Squaring both sides yields the required result.

NOTE The reference to $6.4$ above is simply to what is commonly understood to be the dot product.

$\blacksquare$

Answer 1

Yes, your proof is fine. Why did you write $|\sqrt{a}|$ ? If $a$ is positive, we always have $\sqrt{a}>0$.

Answer 2

it is $$\frac{a}{b}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{d}{c}+\frac{c}{d}+\frac{b}{d}+\frac{d}{b}+4\geq 12+4=16$$ since $$x+\frac{1}{x}\geq 2$$ for $x>0$

Answer 3

While OP's proof is fine, it may be worth noting that the result also follows directly from the AM-HM inequality by simply rewriting the given relation as:

$$ \frac{a+b+c+d}{4} \;\ge\; \frac {4}{\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{c}+\cfrac{1}{d}} $$