sol

Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be positive real numbers such that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=n.$ Prove that$$\frac{a_1+a_2}{a^2_1+a_1a_2+a^2_2}+\frac{a_2+a_3}{a^2_2+a_2a_3+a^2_3}+\cdots+\frac{a_{n-1}+a_n}{a^2_{n-1}+a_{n-1}a_n+a^2_n}+\frac{a_n+a_1}{a^2_n+a_na_1+a^2_1}\leq\frac{2n}{3}.$$

Set $x=1/a$, $y=1/b$, and $z=1/c$. It suffices to show \[ \sum \frac{xy(x+y)}{x^2+xy+y^2}\le 2 \]under $x+y+z=3$. Now, $x^2+xy+y^2\ge 3xy$ by the AM-GM inequality. Thus \[ \frac{xy(x+y)}{x^2+xy+y^2}\le \frac{x+y}{3} \implies \sum \frac{xy(x+y)}{x^2+xy+y^2}\le \frac{2(x+y+z)}{3} =2. \] Remark. My approach destroys immediately the generalization by sqinq.

ALSO DoesPolishing transformations

Simple: $\sum{\frac{a+b}{a^2+ab+b^2}} \le \sum{\frac{a+b}{3ab}} = \frac{2}{3} \sum{\frac{1}{a}} = 2$

Let $a,b,c$ be positive real numbers . Prove that $$\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}\le \frac{2}{\sqrt 3}\sqrt{\frac{1}{ab}+ \frac{1}{bc}+ \frac{1}{ca}}$$ sqing wrote: Let $a_1,a_2,\cdots,a_n (n\ge 3)$ be positive real numbers such that $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=n.$ Prove that$$\frac{a_1+a_2}{a^2_1+a_1a_2+a^2_2}+\frac{a_2+a_3}{a^2_2+a_2a_3+a^2_3}+\cdots+\frac{a_{n-1}+a_n}{a^2_{n-1}+a_{n-1}a_n+a^2_n}+\frac{a_n+a_1}{a^2_n+a_na_1+a^2_1}\leq\frac{2n}{3}.$$

$$\sum_{cyc}\frac{a_1+a_2}{a^2_1+a_1a_2+a^2_2}\leq\sum_{cyc}\frac{a_1+a_2}{3a_1a_2}=\frac{2}{3}\sum_{cyc}\frac{1}{a_1}=\frac{2n}{3}.$$Good.

Attachments:

Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=3$. Prove that $$a+b+c+\frac{2}{a+b+c}\geq \frac{11}{3}$$Solution: $$a+b+c\geq 3$$$$a+b+c+\frac{2}{a+b+c}\geq \frac{7}{9}\cdot 3+\frac{2}{9}\left(a+b+c+\frac{9}{a+b+c}\right)\geq\frac{11}{3}$$

We have $$\sum \frac{a+b}{a^2+ab+b^2} = \sum \frac{\tfrac{1}{a}+\tfrac{1}{b}}{\left(\tfrac{a}{b}+\tfrac{b}{a} \right) + 1} = \sum \frac{3-\tfrac{1}{c}}{\left(\tfrac{a}{b}+\tfrac{b}{a} \right) + 1} \leq \sum \frac{3-\tfrac{1}{c}}{3} = 2$$, where a/b+b/a letting a/b=x>0, and $a/b+b/a=x+1/x\geq2$, so $a/b+b/a+1\geq3$, so denominator is greater or equal to, meaning fraction is less than or equal to. Equality occurs at $(a,b,c)=(1,1,1)$.

$\sum{\frac{a+b}{(a+b)^2-ab}}=\sum{\frac{1}{a+b-\frac{ab}{a+b}}} \leq \sum{\frac{1}{a+b-\frac{a+b}{4}}}=\sum{\frac{4}{3(a+b)}}=\frac{2}{3}.\sum{\frac{2}{a+b}} \leq \frac{2}{3}.\sum{(\frac{1}{2a}+\frac{1}{2b})}=\frac{2}{3}.3=2 $

$$\sum_{cyc}{\frac{a+b}{a^2+ab+b^2}}\leq \sum_{cyc}{\frac{a+b}{3ab}}$$$$=\sum_{cyc}{\frac{a+b}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)ab}}$$$$=\sum_{cyc}{\frac{c(a+b)}{ab+bc+ca}}=2$$

Note, that $\frac{a+b}{a^2+ab+b^2} \le \frac{a+b}{3ab},$ because $a^2+ab+b^2 \ge 3ab,$ because $(a-b)^2 \ge 0,$ then we claim that $\frac{a+b}{3ab}+\frac{a+c}{3ac}+\frac{b+c}{3bc}=\frac{2}{3a}+\frac{2}{3b}+\frac{2}{3c},$ we multiply both sides by $3abc$ so $(a+b)c+(a+c)b+(b+c)a=2ab+2ac+2bc,$ which is the same as the RHS. Now, this is $\frac{2}{3} \cdot \left(\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}\right)=2,$ equality holds when $a=b=c,$ so $a,b,c=(1,1,1).$

My solutions is also similar to others By $AM-GM$ we have: $a^2+b^2 \ge 2 \sqrt{a^2b^2}=2ab$ $=>$ $a^2+b^2 \ge 2ab$ $=>$ $\frac{1}{a^2+b^2}$ $\le$ $\frac{1}{2ab}$ $...(1)$ $\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}=\sum \frac{a+b}{a^2+ab+b^2}$ Combining this with $(1)$ we get: $\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}=\sum \frac{a+b}{a^2+ab+b^2} \le \sum \frac{a+b}{2ab+ab}=\sum \frac{a+b}{3ab}=\frac{a+b}{3ab}+\frac{b+c}{3bc}+\frac{c+a}{3ca}$ $=$$\frac{a}{3ab}+\frac{b}{3ab}+\frac{b}{3bc}+\frac{c}{3bc}+\frac{c}{3ca}+\frac{a}{3ca}=\frac{1}{3b}+\frac{1}{3a}+\frac{1}{3c}+\frac{1}{3b}+\frac{1}{3a}+\frac{1}{3c}$ $=$$2*\frac{1}{3a}+2*\frac{1}{3b}+2*\frac{1}{3c}=\frac{2}{3}*\frac{1}{a}+\frac{2}{3}*\frac{1}{b}+\frac{2}{3}*\frac{1}{c}$ $=$$\frac{2}{3}*\Bigl(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Bigl)=\frac{2}{3}*3=2$ $=>$ $\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2} \le 2$

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Easier than I expected