$\sum \frac{a}{a^2b+2}=\sum \frac{1}{\frac{1}{c}+\frac{2}{a}} \leq \frac{1}{9} \sum (2a+c) =\frac{a+b+c}{3}$

Medjl wrote: $\sum \frac{a}{a^2b+2}=\sum \frac{1}{\frac{1}{c}+\frac{2}{a}} \leq \frac{1}{9} \sum (2a+c) =\frac{a+b+c}{3}$ Nice.

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ \[\frac{{\sum a }}{3} - \sum {\frac{a}{{{a^2}b + 2}}} = \frac{1}{2}\sum {\frac{{{a^3}b}}{{{a^2}b + 2}}} - \frac{{\sum a }}{6} = \sum {\frac{{{a^2}}}{{a + 2c}}} - \frac{{\sum a }}{6} \ge \frac{1}{2}\frac{{{{\left( {\sum a } \right)}^2}}}{{\sum a + 2\sum c }} - \frac{{\sum a }}{6} = 0\]

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}.$$

sqing wrote: Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}.$$ $$\frac{a_1}{a^2_1 a_2\cdots a_{n-1}+n-1} = \frac{1}{\frac{1}{a_n}+\frac{n-1}{a_1}} \leq \frac{(n-1)a_1+a_n}{n^2}, ... $$

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Let $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$.Then the inequality is equivalent to $\frac{x^2z+y^2x+z^2y}{3xyz}\geq \sum_{cyc}\frac{xz}{x^2+2yz}$ Clearing the denominator, we get $2(\sum_{cyc}x^6yz^2)+(\sum_{cyc}x^5z^4)\geq 3(\sum_{cyc}x^4y^2z^3)$ By AM-GM inequality, $2x^6yz^2+z^5y^4\geq 3x^4y^2z^3$ Thus we're done

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Proof of Xiongchangjin：

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Very nice proofs.

adityaguharoy wrote: Very nice proofs. Yeah.

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ $$ \sum_{cyc} \dfrac {a} {a^2 b+2} \le \dfrac {a+b+c} {3} \Leftrightarrow $$$$ \sum_{cyc} \dfrac {c} {a^2 b+2} \ge \dfrac {a+b+c} {3} $$Using Cauchy-Shwarz inequality $$ \sum_{cyc} \dfrac {c} {a^2 b+2} \ge \dfrac {(a+b+c)^2} {a^2 bc+ab^2 c+abc^2 +2(a+b+c)}=\dfrac {a+b+c} {3} $$

Booldy wrote: Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ $$ \sum_{cyc} \dfrac {a} {a^2 b+2} \le \dfrac {a+b+c} {3} \Leftrightarrow $$$$ \sum_{cyc} \dfrac {c} {a^2 b+2} \ge \dfrac {a+b+c} {3} $$Using Cauchy-Shwarz inequality $$ \sum_{cyc} \dfrac {c} {a^2 b+2} \ge \dfrac {(a+b+c)^2} {a^2 bc+ab^2 c+abc^2 +2(a+b+c)}=\dfrac {a+b+c} {3} $$ $$ \sum_{cyc} \dfrac {a} {a^2 b+2} \le \dfrac {a+b+c} {3} \Leftrightarrow \sum_{cyc} \dfrac {c} {a^2 b+2} \ge \dfrac {a+b+c} {3} $$Good.

sqing wrote: Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}.$$ This is very nice.

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Proof of Zhangyanzong: By C-S,

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Very beautiful proof sir. Congratulations !

Let $a$, $b$ and $c$ be positive real numbers satisfying $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \le 3$ . Prove that the following inequality holds: \[\frac{{{a^2}}}{{2{a^2}b + 1}} + \frac{{{b^2}}}{{2{b^2}c + 1}} + \frac{{{c^2}}}{{2{c^2}a + 1}} \le \frac{{a + b + c}}{3}\]http://blog.sina.com.cn/s/blog_5618e6650102xuh9.html

adityaguharoy wrote: Very beautiful proof sir. Congratulations ! Thanks. Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a}{2b+c^2}+\frac{b}{2c+a^2}+\frac{c}{2a+b^2}\le \frac{a^2+b^2+c^2}3.$$

sqing wrote: adityaguharoy wrote: Very beautiful proof sir. Congratulations ! Thanks. Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a}{2b+c^2}+\frac{b}{2c+a^2}+\frac{c}{2a+b^2}\le \frac{a^2+b^2+c^2}3.$$ The same problem was actually also pointed out to me by some of the contestants after the test, particularly the ones who used AM-GM to solve it.

That is very nice. Congratulations !!

sqing wrote: Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Proof of Zhangyanzong: By C-S, How did the first step become the second step by C-S? Attachment: http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOS85L2MzZWJkOWM4N2FlOTBhMzJiOWUyNmYxNDVmY2I2MjM2ZTE2NGE1LmpwZw==&rn=c3FpbmcuLmpwZw==

ENC wrote: sqing wrote: Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Proof of Zhangyanzong: By C-S, How did the first step become the second step by C-S? Attachment: http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvOS85L2MzZWJkOWM4N2FlOTBhMzJiOWUyNmYxNDVmY2I2MjM2ZTE2NGE1LmpwZw==&rn=c3FpbmcuLmpwZw== $$(a^2b+1+1)(\frac{1}{a^2b}+1+1)\geq 9$$

Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}\geq\frac{9}{(a+b+c)^2} .$$(Tiantianxiangshang_ZRH)

sqing wrote: Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}.$$ Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}\geq\frac{n^2}{(a_1+a_2+\cdots + a_n)^2}.$$

Can't we assume $a$ Max {a, b, c} or $a\ge b, c$ then we are Dr one .is it correct!!

Medjl wrote: $\sum \frac{a}{a^2b+2}=\sum \frac{1}{\frac{1}{c}+\frac{2}{a}} \leq \frac{1}{9} \sum (2a+c) =\frac{a+b+c}{3}$ That's my solution. Mr Medjl used AM-HM inequality. sqing wrote: adityaguharoy wrote: Very beautiful proof sir. Congratulations ! Thanks. Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a}{2b+c^2}+\frac{b}{2c+a^2}+\frac{c}{2a+b^2}\le \frac{a^2+b^2+c^2}3.$$ I propose this one. Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$ $\frac{a^2+b^2+c^2}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$

sqing wrote: Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}\geq\frac{n^2}{(a_1+a_2+\cdots + a_n)^2}.$$ Left prove. By AM-HM $\sum_{cyc}\frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1}=\sum_{cyc}\frac{a_1}{\frac{a_1}{a_n}+n-1}\le \sum_{cyc}\left(\frac{a_1}{4}\cdot \frac{a_n}{a_1}+(n-1)\frac{a_1}{4}\right)=\sum_{cyc}\left(\frac{(n-1)a_1+a_n}{4}\right)=\frac{a_1+a_2+\cdots + a_n}{n}$ sqing wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}\geq\frac{9}{(a+b+c)^2} .$$(Tiantianxiangshang_ZRH) By C-S $\sum_{cyc}\frac{a}{a^2b+2}=\sum_{cyc}\frac{1}{ab+2bc}\ge\frac{9}{3(ab+bc+ca)}\ge\frac{9}{(a+b+c)^2} $

sqing wrote: Morskow wrote: Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$ Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{n}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}.$$ Let $a$, $b$ and $c$ be positive real numbers satisfying $abc=1$. Prove that the following inequality holds: $$\sqrt{\frac{ab+bc+ca}{3}}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$$Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. For $\lambda>0,$ prove that the following inequality holds: $$\frac{a_1+a_2+\cdots + a_n}{1+\lambda}\geq \frac{a_1}{a^2_1a_2\cdots a_{n-1}+\lambda} +\frac{a_2}{a^2_2a_3\cdots a_{n}+\lambda}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+\lambda}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+\lambda}.$$

sqing wrote: Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1a_2\cdots a_n=1$. Prove that the following inequality holds: $$\frac{a_1}{a^2_1a_2\cdots a_{n-1}+n-1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+n-1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+n-1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+n-1}\leq \frac{a_1+a_2+\cdots + a_n}{n}.$$ Let $a_1,a_2,\cdots,a_n>0 (n\ge 3)$ and $a_1+a_2+\cdots + a_n=n.$ Prove that the following inequality holds: $$ \frac{a_1}{a^2_1a_2\cdots a_{n-1}+1} +\frac{a_2}{a^2_2a_3\cdots a_{n}+1}+\frac{a_3}{a^2_3a_4\cdots a_{n}a_1+1}+\cdots+\frac{a_n}{a^2_na_1\cdots a_{n-2}+1}\leq\frac{n}{2a_1a_2\cdots a_n}.$$