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Let $p_1, p_2, p_3,.....p_n$ be an arbitary arrangement of natural numbers from $1$ to $2014$. Prove that $$\frac{1}{p_1+p_2} + \frac{1}{p_2+p_3} + \frac{1}{p_3+p_4} + ... + \frac{1}{p_{2013}+p_{2014}} \geq \frac{2013}{2016}$$

EDIT:

Thanks for your responses. Actually, I've already proved the inequality using AM >= HM but I wanted to try a different method so if you could help me out with it, that would be great.

In order to prove the inequality we can say that if all the terms of the sequence (i.e. $\frac{1}{p_1+p_2} + \frac{1}{p_2+p_3} + \frac{1}{p_3+p_4} + ... + \frac{1}{p_{2013}+p_{2014}}$) are minimum and yet their value exceeds $\frac{2013}{2016}$ then we're done.

So what I did was I assigned the values of $p_1, p_2, p_3,.....p_n$ such that their summation gives minimum values of the fractions (which will be when $p_1$ = 1, $p_2$ = 2014, $p_3$ = 2, $p_4$ = 2013 and so on).

Therefore the L.H.S. becomes $\frac{1}{2015} + \frac{1}{2016} + \frac{1}{2015} + ... + \frac{1}{2015}$ which is equal to $\frac{1007}{2015} + \frac{1006}{2016}$.

Subtracting $\frac{1006}{2016}$ from R.H.S., we're left with $\frac{1007}{2015} \geq \frac{1007}{2016}$ which is obviously true.

That's about it. Can anybody tell me if there is anything wrong with this proof?

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    $\begingroup$ If you want to think yourself about it, consider the harmonic mean inequality. $\endgroup$
    – GEO
    Commented Nov 23, 2014 at 16:19

2 Answers 2

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By Cauchy-Schwarz inequality $$\sum_{cyc} \frac1{p_1+p_2} \ge \frac{2013^2}{2014\cdot2015-(p_1+p_{2014})}$$

Hence it is enough to show $$2013 \cdot 2016 \ge 2014 \cdot 2015 - (p_1+p_{2014}) \iff p_1+p_{2014} \ge 2$$

which is obvious.

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We try to proof the following equivalent form: $$\frac{2014}{\frac{1}{p_1+p_2} + \frac{1}{p_2+p_3} + \frac{1}{p_3+p_4} + ... + \frac{1}{p_{2013}+p_{2014}}} \leq \frac{2014 \cdot 2016}{2013}$$ The left side is the harmonic mean which is less or equal the arithmetic one. $$\frac{2014}{\frac{1}{p_1+p_2} + \frac{1}{p_2+p_3} + \frac{1}{p_3+p_4} + ... + \frac{1}{p_{2013}+p_{2014}}}\leq \frac{(p_1+p_2)+(p_2+p_3) + ... + (p_{2013}+p_{2014})}{2014} \leq 2 \cdot \frac{\sum_{i=1}^{2014}p_i}{2014}\leq2015\leq\frac{2014}{2013}\cdot2016$$

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    $\begingroup$ Aren't those the harmonic and arithmetic means of $2013$ numbers? And you don't quite have $2\sum p_i$ numbers in the numerator either. $\endgroup$
    – Nemo
    Commented Nov 23, 2014 at 19:20
  • $\begingroup$ Oops, that is true, thanks for notification. Will correct asap, when I am online on my laptop again. $\endgroup$
    – GEO
    Commented Nov 23, 2014 at 19:38

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