try $\frac{1}{a+b}+\frac{1}{a+c}\ge \sqrt{\frac{2}{a^2+bc}}$(sorry,it's equivalent to $2a^4+b^3c+bc^3\ge a^2b^2+a^2c^2+2a^2bc$ not true )

yefangzhou wrote: Why edit it? $R.H.S^2\le 3\sum_{}^{}\frac{1}{2a^2+2bc}\le 3\sum_{}^{}\frac{1}{2a^2+b^2+c^2}\le \frac{3}{2}\sum_{}^{}\frac{1}{a^2+b^2}$ Are you sure that $3\sum_{}^{}\frac{1}{2a^2+2bc}\le 3\sum_{}^{}\frac{1}{2a^2+b^2+c^2}$ works?

........

this ? $\frac{4}{2a+b+c}\ge \frac{2}{a+\sqrt{bc}}$

yefangzhou wrote: try $\frac{1}{a+b}+\frac{1}{a+c}\ge \sqrt{\frac{2}{a^2+bc}}$(sorry,it's equivalent to $2a^4+b^3c+bc^3\ge a^2b^2+a^2c^2+2a^2bc$ not true ) Not problems! you tried ! But keep trying with this:Without loss of generality, assume that $abc = 1$and $a = min(a; b; c)$. From abc = 1, We get $a \leq{1}$:result $ \frac{1}{a+b}+\frac{1}{a+c}\geq{\frac{2}{a+\sqrt{bc}}}$

$ \frac{1}{a+b}+\frac{1}{a+c}\geq{\frac{2}{a+\sqrt{bc}}}\leftrightarrow (2a+b+c)(a+\sqrt{bc})\ge 2(a+b)(a+c)\leftrightarrow (\sqrt{bc}-a)(\sqrt{b}-\sqrt{c})^2\ge 0\leftrightarrow \sqrt{bc}-a\ge 0$

Dr.frankenstein wrote: Let $a,b,c>0.$ Prove that $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{1}{\sqrt{2a^2+2bc}}+\frac{1}{\sqrt{2b^2+2ca}}+\frac{1}{\sqrt{2c^2+2ab}}$ One auther way is with Cauchy-Schwarz:(previously multiply with $\sqrt{2}$) $(\sum_{cyc}\frac{1}{\sqrt{a^2+bc}})^2\leq{(\sum_{cyc}\frac{1}{(a+b)(a+c)})(\sum_{cyc}\frac{(a+b)(a+c)}{a^2+bc})}=$ $=\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3] $ .Is suffices to prove that $\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3]\leq{2(\sum_{cyc}\frac{1}{a+b})^2} $ Or ,equivalent $ \sum_{cyc}\frac{a(b+c)}{a^2+bc}-3\leq{\frac{\sum_{cyc}a^4-\sum_{cyc}a^2b^2}{(a+b)(b+c)(c+a)\sum_{cyc}a}}$ or $ \sum_{cyc}(a-b)(a-c)(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq0$.Due symetry ,we may assume ,wtith you, $a\geq{b}\geq{c} $ since $ a-c\geq{\frac{a}{b}(b-c)}$ ,than is suffices to prove that $a(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq{b(\frac{1}{b^2+ca}+\frac{1}{(a+c)(a+b+c)})}$ OR $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ wich is true!. Equality holds if and only if $ a=b=c $.

I have an idea: assume that $a\ge b\ge c$ $R.H.S^2\le \sum_{}^{}\frac{3}{2a^2+2bc}\le \frac{3}{(a+b)(a+c)}+\frac{3}{(c+b)(c+a)}+\frac{3}{2b^2+2ac}$

yefangzhou wrote: $ \frac{1}{a+b}+\frac{1}{a+c}\geq{\frac{2}{a+\sqrt{bc}}}\leftrightarrow (2a+b+c)(a+\sqrt{bc})\ge 2(a+b)(a+c)\leftrightarrow (\sqrt{bc}-a)(\sqrt{b}-\sqrt{c})^2\ge 0\leftrightarrow \sqrt{bc}-a\ge 0$ ??? (not true I think) is true because $a = min(a; b; c)$. From abc = 1, We get $a \leq{1}$:

Redacted ...

BestChoice123 wrote: teomihai wrote: yefangzhou wrote: $ \frac{1}{a+b}+\frac{1}{a+c}\geq{\frac{2}{a+\sqrt{bc}}}\leftrightarrow (2a+b+c)(a+\sqrt{bc})\ge 2(a+b)(a+c)\leftrightarrow (\sqrt{bc}-a)(\sqrt{b}-\sqrt{c})^2\ge 0\leftrightarrow \sqrt{bc}-a\ge 0$ ??? (not true I think) is true because $a = min(a; b; c)$. From abc = 1, We get $a \leq{1}$: What about $b$ or $c$? $ bc=\frac{1}{a} $ see 8- ieme post please!

teomihai wrote: Dr.frankenstein wrote: Let $a,b,c>0.$ Prove that $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{1}{\sqrt{2a^2+2bc}}+\frac{1}{\sqrt{2b^2+2ca}}+\frac{1}{\sqrt{2c^2+2ab}}$ One auther way is with Cauchy-Schwarz:(previously multiply with $\sqrt{2}$) $(\sum_{cyc}\frac{1}{\sqrt{a^2+bc}})^2\leq{(\sum_{cyc}\frac{1}{(a+b)(a+c)})(\sum_{cyc}\frac{(a+b)(a+c)}{a^2+bc})}=$ $=\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3] $ .Is suffices to prove that $\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3]\leq{2(\sum_{cyc}\frac{1}{a+b})^2} $ Or ,equivalent $ \sum_{cyc}\frac{a(b+c)}{a^2+bc}-3\leq{\frac{\sum_{cyc}a^4-\sum_{cyc}a^2b^2}{(a+b)(b+c)(c+a)\sum_{cyc}a}}$ or $ \sum_{cyc}(a-b)(a-c)(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq0$.Due symetry ,we may assume ,wtith you, $a\geq{b}\geq{c} $ since $ a-c\geq{\frac{a}{b}(b-c)}$ ,than is suffices to prove that $a(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq{b(\frac{1}{b^2+ca}+\frac{1}{(a+c)(a+b+c)})}$ OR $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ wich is true!. Equality holds if and only if $ a=b=c $. it is O.K.?

teomihai wrote: $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ Maybe this step. The rest is beautiful.

arqady wrote: teomihai wrote: $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ Maybe this step. The rest is beautiful. Thanks,sir ,for this $ \sum_{cyc}\frac{a(b+c)}{a^2+bc}-3\leq{\frac{\sum_{cyc}a^4-\sum_{cyc}a^2b^2}{(a+b)(b+c)(c+a)\sum_{cyc}a}}$ because $\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3]\leq{2(\sum_{cyc}\frac{1}{a+b})^2} $ is equivalent with $ \sum_{cyc}\frac{a(b+c)}{a^2+bc}+3\leq{\frac{(\sum_{cyc}a^2+3\sum_{cyc}ab)^2}{(a+b)(b+c)(c+a)\sum_{cyc}a}}$ and this step $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ ,because $a(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq{b(\frac{1}{b^2+ca}+\frac{1}{(a+c)(a+b+c)})}$ is equivalent (after denominateur) with $a^4 c + a^3 (c^2 - b c) + a^2 (b c^2 + c^3) + a (b^3 c - b^2 c^2) - b^2 c (b^2 + b c + c^2)\geq0 $ or ,decomposing into factors,we have $c (a - b) (a + b) (a^2 - a b + a c + b^2 + b c + c^2)\geq0$ ,or $c(a^2-b^2)([(a-b)^2+ab+bc+ca+c^2]\geq0$, done with due symetry ,we may assume ,wtith you (yefangzhou), $a\geq{b}\geq{c} $ since $ a-c\geq{\frac{a}{b}(b-c)}$ .

You forgot $c^2$ here: $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$

arqady wrote: You forgot $c^2$ here: $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ Yes,sorry ,i corected.

teomihai wrote: Dr.frankenstein wrote: Let $a,b,c>0.$ Prove that $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{1}{\sqrt{2a^2+2bc}}+\frac{1}{\sqrt{2b^2+2ca}}+\frac{1}{\sqrt{2c^2+2ab}}$ One auther way is with Cauchy-Schwarz:(previously multiply with $\sqrt{2}$) $(\sum_{cyc}\frac{1}{\sqrt{a^2+bc}})^2\leq{(\sum_{cyc}\frac{1}{(a+b)(a+c)})(\sum_{cyc}\frac{(a+b)(a+c)}{a^2+bc})}=$ $=\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3] $ .Is suffices to prove that $\frac{2\sum_{cyc}a}{(a+b)(b+c)(c+a)}[\sum_{cyc}\frac{a(b+c)}{a^2+bc}+3]\leq{2(\sum_{cyc}\frac{1}{a+b})^2} $ ($\ast$) Or ,equivalent $ \sum_{cyc}\frac{a(b+c)}{a^2+bc}-3\leq{\frac{\sum_{cyc}a^4-\sum_{cyc}a^2b^2}{(a+b)(b+c)(c+a)\sum_{cyc}a}}$ or $ \sum_{cyc}(a-b)(a-c)(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq0$.Due symetry ,we may assume ,wtith you, $a\geq{b}\geq{c} $ since $ a-c\geq{\frac{a}{b}(b-c)}$ ,than is suffices to prove that $a(\frac{1}{a^2+bc}+\frac{1}{(b+c)(a+b+c)})\geq{b(\frac{1}{b^2+ca}+\frac{1}{(a+c)(a+b+c)})}$ OR $c(a^2-b^2)([(a-b)^2+ab+bc+ca]\geq0$ wich is true!. Equality holds if and only if $ a=b=c $. How do you found this idea to use C-S$?$ I tried $(\sum \frac{1}{\sqrt{a^2+bc}})^2 \leqq 3(\sum\frac{1}{a^2+bc})$ but no success. Another way to prove $(\ast)$ is use MV method: $f(a;b;c) \geqq f(a;\frac{(b+c)}{2} ;\frac{(b+c)}{2})$ with $a=\min\{a,b,c\}$. Which's very easy!

work and this!