Let $a_1, a_2,...,a_n$ be positive real numbers, prove that $$\sum {\frac{a_{i+1}}{a_i}} \ge \sum{\sqrt{\frac{a_{i+1}^2+1}{a_i^2+1}}}$$$a_{n+1}=a_1$
Problem
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Tags: inequalities, algebra
Daniel2003
28.12.2017 14:09
any ideas?
ManuelKahayon
28.12.2017 16:06
\(\frac{x}{y} - \sqrt{\frac{x^2+1}{y^2+1}} \geq \frac{x}{y} \sqrt{\frac{y^2+1}{x^2+1}} - 1\)
Which can be easily proven through full expansion and squaring both sides after transposing the negative terms. It boils down to \(x^4 + y^4 \geq 2x^2y^2\), which is obviously true. Then the problem becomes a simple AM-GM exercise.
MariusStanean
28.12.2017 20:03
My solution: let $a_i=\tan\alpha_i$ where $\alpha_i\in(0,\pi/2)$. Hence, we need to prove that $$\sum_{i=1}^n\frac{\sin\alpha_{i+1}}{\sin\alpha_i}\cdot\frac{\cos\alpha_i}{\cos\alpha_{i+1}}\ge \sum_{i=1}^n\frac{\cos\alpha_i}{\cos\alpha_{i+1}},$$or $$\sum_{i=1}^n\frac{\cos\alpha_i}{\cos\alpha_{i+1}}\cdot\frac{\sin\alpha_{i+1}-\sin\alpha_i}{\sin\alpha_i}\ge 0,$$or $$\sum_{i=1}^n\frac{\sin\alpha_{i+1}-\sin\alpha_i}{\sin\alpha_i}+\sum_{i=1}^n\frac{\cos\alpha_i-\cos\alpha_{i+1}}{\cos\alpha_{i+1}}\cdot\frac{\sin\alpha_{i+1}-\sin\alpha_i}{\sin\alpha_i}\ge 0$$which is obviously true, first sum from AM-GM Inequality and second sum from monotony.
whiwho
28.12.2017 20:20
ManuelKahayon wrote:
\(\frac{x}{y} - \sqrt{\frac{x^2+1}{y^2+1}} \geq \frac{x}{y} \sqrt{\frac{y^2+1}{x^2+1}} - 1\)
Which can be easily proven through full expansion and squaring both sides after transposing the negative terms. It boils down to \(x^4 + y^2 \geq 2x^2y^2\), which is obviously true. Then the problem becomes a simple AM-GM exercise.
If $x^4+y^2 \geq 2x^2y^2$, then, $1^4+100^2 \geq 2*1^2*100^2 \implies 10001 \geq 20000$
teomihai
28.12.2017 20:33
very nice solution,Marius Stanean Happy new years
ManuelKahayon
29.12.2017 17:53
whiwho wrote: ManuelKahayon wrote:
\(\frac{x}{y} - \sqrt{\frac{x^2+1}{y^2+1}} \geq \frac{x}{y} \sqrt{\frac{y^2+1}{x^2+1}} - 1\)
Which can be easily proven through full expansion and squaring both sides after transposing the negative terms. It boils down to \(x^4 + y^2 \geq 2x^2y^2\), which is obviously true. Then the problem becomes a simple AM-GM exercise.
If $x^4+y^2 \geq 2x^2y^2$, then, $1^4+100^2 \geq 2*1^2*100^2 \implies 10001 \geq 20000$ Uhh, I am so sorry. This has been fixed now.