https://artofproblemsolving.com/community/c6t243f6h1486673_seems_hard

sqing wrote: Let $a_1,a_2,\cdots,a_{n+1}>0$. Prove that$$\sum_{i-1}^{n}a_i\sum_{i=1}^{n}a_{i+1}\geq \sum_{i=1}^{n}\frac{a_i a_{i+1}}{a_i+a_{i+1}}\cdot \sum_{i=1}^{n}(a_i+a_{i+1})$$

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})}$

how to prove the first inequality? $\frac{ab}{a+b}* \frac{bc}{b+c} \le \frac{(a+b)(b+c)}{(a+b)+(b+c)}$

Mathskidd wrote: how to prove the first inequality? $\frac{ab}{a+b}+ \frac{bc}{b+c} \le \frac{(a+b)(b+c)}{(a+b)+(b+c)}$ $\iff \frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{b}+\frac{1}{c}}\leq \frac{1}{\frac{1}{a+b}+\frac{1}{b+c}}$

ab/(a+b)+bc/(b+c)

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.

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 \[ \frac{(a+b)(b+c)}{(a+b)+(b+c)} + \frac{cd}{c+d} \le \frac{(a+b+c)(b+c+d)}{(a+b+c)+(b+c+d)} \]

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*}

I still wonder if there is another elegant way to see the validity of the lemma above.

https://artofproblemsolving.com/community/c6t243f6h1527596_convex_function_inequlity

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.

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)=...

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