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Prove that $$\frac{1}{a_1} + \frac{2}{a_1+a_2} + \frac{3}{a_1+a_2+a_3}<2(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}) $$ where $a_1, a_2, a_3 >0$.
From AM-HM I got that $\frac{2}{a_1+a_2}\le \frac{1}{2}(\frac{1}{a_1}+\frac{1}{a_2})$ and $\frac{3}{a_1+a_2+a_3}\le \frac{1}{3}(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3})$, but adding these is not enough.

ChemistryGeek
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    Unless I am mistaken, adding these inequalities is good enough. You get $\frac{11}{6} \frac{1}{a_1} + \frac{5}{6} \frac{1}{a_2} + \frac{1}{3} \frac{1}{a_3}$ – Martin R Apr 23 '20 at 12:11
  • It is good enough, indeed! $1+\frac{1}{2}+\frac{1}{3} < 2$. – rtybase Apr 23 '20 at 12:13

2 Answers2

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Further to my comment and your progress with AM-HM $$\frac{1}{a_1}<\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}$$ $$\frac{2}{a_1+a_2}\leq \frac{1}{2}\left(\frac{1}{a_1}+\frac{1}{a_2}\right)<\frac{1}{2}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)$$ $$\frac{3}{a_1+a_2+a_3}\leq \frac{1}{3}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)$$ Add together $$\frac{1}{a_1}+\frac{2}{a_1+a_2}+\frac{3}{a_1+a_2+a_3}<\color{red}{\left(1+\frac{1}{2}+\frac{1}{3}\right)}\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)<\color{red}{2}\cdot ...$$

rtybase
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Adding the inequalities given by you: $$\dfrac{1}{a_1}+\dfrac{2}{a_1+a_2}+\dfrac{3}{a_1+a_2+a_3}<\dfrac{1+\frac{1}{2}+\frac{1}{3}}{a_1}+\dfrac{\frac{1}{2}+\frac{1}{3}}{a_2}+\dfrac{\frac{1}{3}}{a_3}$$ $$\dfrac{1+\frac{1}{2}+\frac{1}{3}}{a_1}+\dfrac{\frac{1}{2}+\frac{1}{3}}{a_2}+\dfrac{\frac{1}{3}}{a_3}<\dfrac{2}{a_1}+\dfrac{2}{a_2}+\dfrac{2}{a_3}$$ $\therefore \dfrac{1}{a_1}+\dfrac{2}{a_1+a_2}+\dfrac{3}{a_1+a_2+a_3}<\dfrac{2}{a_1}+\dfrac{2}{a_2}+\dfrac{2}{a_3}$

Devansh Kamra
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  • I must admit, you re-grouping is right, but a bit "asking for more concentration", thus, I posted my answer ... +1 from me anyway. – rtybase Apr 23 '20 at 12:37