Let $a_1,a_2,\cdots,a_{n+1}>0$. Prove that$$\sum_{i-1}^{n}a_i\sum_{i=1}^{n}a_{i+1}\geq \sum_{i=1}^{n}\frac{a_i a_{i+1}}{a_i+a_{i+1}}\cdot \sum_{i=1}^{n}(a_i+a_{i+1})$$
Problem
Source: China Jiangxi , Jul 30, 2017
Tags: inequalities, algebra, China, BPSQ
30.07.2017 09:29
https://artofproblemsolving.com/community/c6t243f6h1486673_seems_hard
30.07.2017 12:40
sqing wrote: Let $a_1,a_2,\cdots,a_{n+1}>0$. Prove that$$\sum_{i-1}^{n}a_i\sum_{i=1}^{n}a_{i+1}\geq \sum_{i=1}^{n}\frac{a_i a_{i+1}}{a_i+a_{i+1}}\cdot \sum_{i=1}^{n}(a_i+a_{i+1})$$
30.07.2017 16:37
sqing wrote: Let $a_1,a_2,\cdots,a_{n+1}>0$. Prove that$$\sum_{i-1}^{n}a_i\sum_{i=1}^{n}a_{i+1}\geq \sum_{i=1}^{n}\frac{a_i a_{i+1}}{a_i+a_{i+1}}\cdot \sum_{i=1}^{n}(a_i+a_{i+1})$$ CeuAzul wrote: Solution Since $\frac{ab}{a+b}* \frac{bc}{b+c} \le \frac{(a+b)(b+c)}{(a+b)+(b+c)}$ We have $ \sum_{i=1}^{n} \frac{a_ia_{i+1}}{a_i+a_{i+1}} \le \frac{(a_1+a_2)(a_2+a_3)}{(a_1+a_2)+(a_2+a_3)}+ \sum_{i=3}^{n} \frac{a_ia_{i+1}}{a_i+a_{i+1}} \le \cdots \le \frac{\sum_{i=1}^{n}a_i \sum_{i=1}^{n}a_{i+1}}{\sum_{i=1}^{n} (a_i+a_{i+1})}$
30.07.2017 16:41
how to prove the first inequality? $\frac{ab}{a+b}* \frac{bc}{b+c} \le \frac{(a+b)(b+c)}{(a+b)+(b+c)}$
30.07.2017 17:01
Mathskidd wrote: how to prove the first inequality? $\frac{ab}{a+b}+ \frac{bc}{b+c} \le \frac{(a+b)(b+c)}{(a+b)+(b+c)}$ $\iff \frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}\leq \frac{1}{\frac{1}{a+b}+\frac{1}{b+c}}$
30.07.2017 17:07
ab/(a+b)+bc/(b+c)
05.09.2017 20:25
Mathskidd wrote: how to prove the first inequality? My answer is the following. $\frac{ab}{a+b} + \frac{bc}{b+c} = b \left(\frac{a}{a+b} + \frac{c}{b+c} \right)$ $= b \left( 2- \frac{b}{a+b}-\frac{b}{b+c} \right) \ge \frac{b}{\frac{b}{a+b} + \frac{b}{b+c}}$ $= \frac{(a+b)(b+c)}{a+2b+c}$ However, I wonder if there is another marvelous point of view so that the inequality becomes trivial.
12.10.2017 11:34
Another qutestion; How do we apply $\frac{ab}{a+b} + \frac{bc}{b+c} \le \frac{(a+b)(b+c)}{(a+b)+(b+c)}$ to prove \[ \frac{(a+b)(b+c)}{(a+b)+(b+c)} + \frac{cd}{c+d} \le \frac{(a+b+c)(b+c+d)}{(a+b+c)+(b+c+d)} \]
12.10.2017 11:44
Lemma: For every positive $a,b,c,d$, we have \[ \frac{ab}{a+b} + \frac{cd}{c+d} \le \frac{(a+c)(b+d)}{a+b+c+d} \]<proof> Multiplying both sides by $(a+b)(c+d)(a+b+c+d)$, we have the equivalent \[ (bc-ad)^2 \ge 0 \]∎ Applying lemma we have for positive $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$, \begin{align*} & \dfrac{a_1a_2}{a_1 + a_2} + \dfrac{a_2a_3}{a_2+a_3} + \cdots \\ &\le \dfrac{(a_1+a_2)(a_2+a_3)}{a_1+a_2+a_2+a_3} + \dfrac{a_3 a_4}{a_3+a_4} + \cdots \\ &\le \dfrac{(a_1+a_2+a_3)(a_2+a_3+a_4)}{a_1+a_2+a_3+a_2+a_3+a_4} + \dfrac{a_4 a_5}{a_4+a_5} + \cdots \\ & \vdots \\ & \le \dfrac{(a_1+a_2+ \cdots + a_n)(a_2+a_3+ \cdots +a_{n+1})}{a_1+a_2+ \cdots + a_n +a_2+a_3+ \cdots +a_{n+1}} \end{align*}
12.10.2017 11:48
I still wonder if there is another elegant way to see the validity of the lemma above.
12.10.2017 12:47
https://artofproblemsolving.com/community/c6t243f6h1527596_convex_function_inequlity
12.10.2017 18:15
There is a simple way to prove the lemma. We have an identity $\frac{4xy}{x+y}=x+y-\frac{(x-y)^2}{x+y}$ Hence, the lemma is equivalent to $\frac{(a-b)^2}{a+b}+\frac{(c-d)^2}{c+d}\geq\frac{(a+c-b-d)^2}{a+b+c+d}$ which is an immediate application of C-S.
21.11.2018 11:06
Of course there's a very elegant way. The ab/(a+b) form is a strong hint to resistors connected in parallel. The total resistance of the circuit below is obviously smaller or equal to the circuit above, the equality holds iff all the bridges are already balanced, i.e. a(1):a(2)=a(2):a(3)=...
Attachments:
