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Use the Cauchy-Schwarz inequality on the euclidean space $ \mathbb{R}^3 $ (usual inner product) to show that, given 3 strictly positive numbers $a_1, a_2, a_3$ we have

$$ (a_1 + a_2 + a_3) \left(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}\right) \geq 9 $$


My attempt was to expand this inequality and check if I could find 2 vectors of $\mathbb{R}^3 $ to prove the inequality. We have:

$ 1 + 1 + 1 + \frac{a_1}{a_2} + \frac{a_1}{a_3} + \frac{a_2}{a_1} + \frac{a_2}{a_3} + \frac{a_3}{a_1} + \frac{a_3}{a_2} \geq 9 \Leftrightarrow$

$a_1^2a_2 + a_1^2 a_3 + a_2^2 a_1 + a_2^2 a_3 + a_3^2 a_1 + a_3^2 a_2 \geq 6 a_1 a_2 a_3 \Leftrightarrow$

$ (a_1 + a_2)^3 + (a_1+a_3)^3 + (a_2+a_3)^3 \geq 2a_1(a_1^2 + a_2a_3) + 2a_2(a_2^2 + a_1a_3) + 2a_3(a_3^2 + a_1a_2)$

so I tried $v=(2a_1, 2a_2, 2a_3) $ and $u=(a_1^2 + a_2a_3, a_2^2 + a_1a_3, a_3^2 + a_1a_2)$ but I had no sucess (well, I couldn't verify the left side of the inequality). Maybe a should put the coefficients not like this

Is there a better way to solve it? Maybe using a consequence of the cauchy-schwarz inequality

Thanks for the help!

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If you want it using Cauchy-Schwarz, set $b_i=\sqrt{a_i}$. Then CS tells you that $$\left(\sum (\frac{1}{b_i})^2\right)\left(\sum b_i^2\right)\ge \left(\sum b_i\frac{1}{b_i}\right)^2=3^2=9$$

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Hint: $\dfrac{a_1}{a_2}+\dfrac{a_2}{a_1}\geqslant 2$.

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  • $\begingroup$ Got it! Thanks for the help $\endgroup$ – Giiovanna Oct 7 '13 at 18:09

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