$A_{max}=2.$

Posted: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=town&t=24768

Romania Team Selection Test 1993: Find the greatest real numbers $A$ such that $\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}> A$ is true for all positve reals numberes $x,y,z$ . Osh town olympiad: For $a,b$ and $c$ positive reals prove that\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2.\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=293319#p293319 Macedonia,1995

For $a,b$ and $c$ positive reals prove that\[\sqrt{\frac{a}{b+c}}+2\sqrt{\frac{b}{c+a}}+4\sqrt{\frac{c}{a+b}}\leq \sqrt{7\left(\frac{a}{b+c}+\frac{2b}{c+a}+\frac{4c}{a+b}\right)} .\]\[\sqrt{\frac{a}{b}}+2\sqrt{\frac{b}{c}}+3\sqrt{\frac{c}{a}}\leq \sqrt{6\left(\frac{a}{b}+\frac{2b}{c}+\frac{3c}{a}\right)} .\]( Daniel Sitaru)

Let $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ be positive real numbers. Prove that$$\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\leq \sqrt{(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)}.$$$$\sqrt{\frac{a_1}{a_2}}+2\sqrt{\frac{a_2}{a_3}}+\cdots+(n-1)\sqrt{\frac{a_{n-1}}{a_n}}+n\sqrt{\frac{a_n}{a_1}}\leq \sqrt{\frac{n(n+1)}{2}\left(\frac{a_1}{a_2}+\frac{2a_2}{a_3}+\cdots+\frac{(n-1)a_{n-1}}{a_n}+\frac{na_n}{a_1}\right)}.$$

sqing wrote: Let $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ be positive real numbers. Prove that$$\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\leq \sqrt{(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)}.$$$$\sqrt{\frac{a_1}{a_2}}+2\sqrt{\frac{a_2}{a_3}}+\cdots+(n-1)\sqrt{\frac{a_{n-1}}{a_n}}+n\sqrt{\frac{a_n}{a_1}}\leq \sqrt{\frac{n(n+1)}{2}\left(\frac{a_1}{a_2}+\frac{2a_2}{a_3}+\cdots+\frac{(n-1)a_{n-1}}{a_n}+\frac{na_n}{a_1}\right)}.$$ Solution. Since that $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ are positive real numbers, the Cauchy-Schwarz's inequality gives that \begin{align*}&(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)\\ =&\left(1+2+\cdots+2^{n-1}\right)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)\\ \ge&\left(\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\right)^2, \end{align*}which yields the desired inequality immediately. The second inequality can be shown likewise. $\blacksquare$

ytChen wrote: sqing wrote: Let $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ be positive real numbers. Prove that$$\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\leq \sqrt{(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)}.$$$$\sqrt{\frac{a_1}{a_2}}+2\sqrt{\frac{a_2}{a_3}}+\cdots+(n-1)\sqrt{\frac{a_{n-1}}{a_n}}+n\sqrt{\frac{a_n}{a_1}}\leq \sqrt{\frac{n(n+1)}{2}\left(\frac{a_1}{a_2}+\frac{2a_2}{a_3}+\cdots+\frac{(n-1)a_{n-1}}{a_n}+\frac{na_n}{a_1}\right)}.$$ Solution. Since that $a_1,a_2,\cdots,a_n $ $(n\ge 2)$ are positive real numbers, the Cauchy-Schwarz's inequality gives that \begin{align*}&(2^n-1)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)\\ =&\left(1+2+\cdots+2^{n-1}\right)\left(\frac{a_1}{a_2+a_3}+\frac{2a_2}{a_3+a_4}+\cdots+\frac{2^{n-2}a_{n-1}}{a_n+a_1}+\frac{2^{n-1}a_n}{a_1+a_2}\right)\\ \ge&\left(\sqrt{\frac{a_1}{a_2+a_3}}+2\sqrt{\frac{a_2}{a_3+a_4}}+\cdots+2^{n-2}\sqrt{\frac{a_{n-1}}{a_n+a_1}}+2^{n-1}\sqrt{\frac{a_n}{a_1+a_2}}\right)^2, \end{align*}which yields the desired inequality immediately. The second inequality can be shown likewise. $\blacksquare$ Thank you very much.