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Using the inequality of Cauchy-Schwarz in Euclidean space $\mathbb{R}^3$(with the usual product), show that if $a,b,c \in \mathbb{R}^+$ different from zero, then:$$9 \leq (a+b+c)\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ Inequality of Cauchy-Schwarz, $|\langle\ v\ ,\ u\ \rangle|\leq||v||\cdot||u||, \forall v,u \in\mathbb{R}^3$.

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Hint Compare the RHS. Can you think of any vectors $u,v$ so that

$$\| u \|^2=a+b+c$$ $$\|v\|^2=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

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