It's equivalent to $x,y,z>0\wedge \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\implies \sqrt{x+y}+\sqrt{y+z}+\sqrt{x+z}\ge 3$

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parmenides51 wrote: Let $a,b,c$ be positive real numbers. Prove that $\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3$ The following can help. https://artofproblemsolving.com/community/c6h574423p3386391 The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\prod_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\geq1.$$

https://mp.weixin.qq.com/s?__biz=MzU0NDI5MjczNw==&mid=2247484855&idx=1&sn=3616b1064fc591ee1c8c5ec64b3a6d3c&chksm=fb7f2f59cc08a64f10c9e2b3340565bbc1f6ff1f44787f7e7d4c0aa909785204e10300211564&mpshare=1&scene=1&srcid=0915P9oSD0WAI5fbE2gSdaLg&pass_ticket=Tmf35tS1jteQMk9IOMIWc3%2Bfof7evQMTqPlz%2FgCIQ%2Fo%3D#rd

$x,y,z>0$,prove that $3\, \left( {\frac {x}{y+z}}+{\frac {y}{z+x}} \right) \left( {\frac {y }{z+x}}+{\frac {z}{x+y}} \right) \left( {\frac {z}{x+y}}+{\frac {x}{y +z}} \right) \geq \sqrt {{\frac {x}{y+z}}+{\frac {y}{z+x}}}+\sqrt {{\frac {y}{z+x}}+{\frac {z}{x+y}}}+\sqrt {{\frac {z}{x+y}}+{\frac {x}{y+z}}} $

arqady wrote: Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\prod_{cyc}\left(\frac{a}{b+c}+\frac{b}{a+c}\right)\geq1.$$ Thank you arqady $\frac{a}{b+c}+\frac{b}{c+a}\ge \frac{a+b}{2}\left(\frac{1}{b+c}+\frac{1}{c+a}\right)\ge \frac{a+b}{\sqrt{(b+c)(c+a)}}$

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parmenides51 wrote: Let $a,b,c$ be positive real numbers. Prove that $\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3$ Let $a,b,c>0 .$ Prove inequality\[ (\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})\ge1.\] sqing wrote: sqing wrote: Let $a,b,c>0 $ ,prove inequality\[ (\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})\ge1.\] $\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{b+c}{2}\left(\frac{1}{c+a}+ \frac{1}{a+b}\right)\geq\frac{b+c}{\sqrt{(c+a)(a+b)}},\cdots$ \[\left( \frac{a}{b+c}+\frac{b}{c+a} \right)\left( \frac{b}{c+a}+\frac{c}{a+b} \right)\left( \frac{c}{a+b}+\frac{a}{b+c} \right)\ge 1+\frac{1}{8}{{\left( \frac{a-b}{a+b} \right)}^{2}}{{\left( \frac{b-c}{b+c} \right)}^{2}}{{\left( \frac{c-a}{c+a} \right)}^{2}}.\]\[\left( \frac{a}{b+c}+\frac{b}{c+a} \right)\left( \frac{b}{c+a}+\frac{c}{a+b} \right)\left( \frac{c}{a+b}+\frac{a}{b+c} \right)\ge\frac{(a+b+c)^2}{3(ab+bc+ca)}\]

${\frac {x}{y+z}}+{\frac {y}{z+x}}+{\frac {z}{x+y}}\geq {\frac {x}{\sqrt { \left( x+y \right) \left( y+z \right) }}}+{\frac {y}{\sqrt { \left( y+z \right) \left( z+x \right) }}}+{\frac {z}{\sqrt { \left( z+x \right) \left( x+y \right) }}} $ In acute triangle, $x=b^2+c^2-a^2...$ But any triangle,have: ${\frac {{b}^{2}+{c}^{2}-{a}^{2}}{{a}^{2}}}+{\frac {-{b}^{2}+{a}^{2}+{c }^{2}}{{b}^{2}}}+{\frac {{b}^{2}+{a}^{2}-{c}^{2}}{{c}^{2}}}\geq {\frac {{b }^{2}+{c}^{2}-{a}^{2}}{ac}}+{\frac {-{b}^{2}+{a}^{2}+{c}^{2}}{ab}}+{ \frac {{b}^{2}+{a}^{2}-{c}^{2}}{bc}} $ Done

parmenides51 wrote: Let $a,b,c$ be positive real numbers. Prove that $\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3$ Very simple inequality. Just use AM-GM.

Sqing, if you use directly AM-GM you have to prove $\frac{abc}{(a+b)(b+c)(c+a)}\ge \frac{1}{8}$ which isn't true. Could you elaborate more on your idea?

In acute triangle,prove \[\frac{1}{2}\,{\frac {{b}^{2}+{c}^{2}-{a}^{2}}{ac}}+\frac{1}{2}\,{\frac {-{b}^{2}+{a}^{ 2}+{c}^{2}}{ab}}+\frac{1}{2}\,{\frac {{b}^{2}+{a}^{2}-{c}^{2}}{bc}}\geq \sin \left( C \right) \sqrt {\cot \left( A \right) \cot \left( B \right) } +\sin \left( A \right) \sqrt {\cot \left( B \right) \cot \left( C \right) }+\sin \left( B \right) \sqrt {\cot \left( C \right) \cot \left( A \right) }\]

sqing wrote: parmenides51 wrote: Let $a,b,c$ be positive real numbers. Prove that $\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3$ Very simple inequality. Just use AM-GM. yes! Very simple inequality.

xzlbq wrote: $x,y,z>0$,prove that $3\, \left( {\frac {x}{y+z}}+{\frac {y}{z+x}} \right) \left( {\frac {y }{z+x}}+{\frac {z}{x+y}} \right) \left( {\frac {z}{x+y}}+{\frac {x}{y +z}} \right) \geq \sqrt {{\frac {x}{y+z}}+{\frac {y}{z+x}}}+\sqrt {{\frac {y}{z+x}}+{\frac {z}{x+y}}}+\sqrt {{\frac {z}{x+y}}+{\frac {x}{y+z}}} $ But this still hard

$x,y,z>0$,prove that $\left( {\frac {x}{y+z}}+{\frac {y}{z+x}} \right) \left( {\frac {y}{z +x}}+{\frac {z}{x+y}} \right) \left( {\frac {z}{x+y}}+{\frac {x}{y+z} } \right) $ $\geq {\frac {27}{8}}\, \left( {\frac {x}{y+z}}+\frac{1}{4}\,{\frac {y+z} {x+y+z}} \right) \left( {\frac {y}{z+x}}+\frac{1}{4}\,{\frac {z+x}{x+y+z}} \right) \left( {\frac {z}{x+y}}+\frac{1}{4}\,{\frac {x+y}{x+y+z}} \right) \geq 1 $

WolfusA wrote: Sqing, if you use directly AM-GM you have to prove $\frac{abc}{(a+b)(b+c)(c+a)}\ge \frac{1}{8}$ which isn't true. Could you elaborate more on your idea? By AM-GM, $$\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3\sqrt[6]{ (\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}\geq3\sqrt[6]{ \frac{b+c}{\sqrt{(c+a)(a+b)}}\cdot \frac{c+a}{\sqrt{(a+b)(b+c)}}\cdot \frac{a+b}{\sqrt{(b+c)(c+a)}}}=3$$

sqing wrote: WolfusA wrote: Sqing, if you use directly AM-GM you have to prove $\frac{abc}{(a+b)(b+c)(c+a)}\ge \frac{1}{8}$ which isn't true. Could you elaborate more on your idea? By AM-GM, $\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3\sqrt[6]{ (\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}$ $\geq3\sqrt[6]{ \frac{b+c}{\sqrt{(c+a)(a+b)}}\cdot \frac{c+a}{\sqrt{(a+b)(b+c)}}\cdot \frac{a+b}{\sqrt{(b+c)(c+a)}}}=3$ Nice!

\[2+\sqrt {2}>\sqrt {{\frac {x}{x+y}}+{\frac {y}{y+z}}}+\sqrt {{\frac {y }{y+z}}+{\frac {z}{z+x}}}+\sqrt {{\frac {z}{z+x}}+{\frac {x}{x+y}}} \geq 2+8\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }} \] \[\sqrt {{\frac {x}{x+y}}+{\frac {y}{y+z}}}+\sqrt {{\frac {y}{y+z}}+{ \frac {z}{z+x}}}+\sqrt {{\frac {z}{z+x}}+{\frac {x}{x+y}}}\geq 2+2\,\sqrt {2}\sqrt {{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }}} \]

$a,b,c,d>0$,prove that \[{\frac {4}{27}}\geq {\frac {bcd}{ \left( a+c+d \right) \left( d+a+b \right) \left( a+b+c \right) }}+{\frac {cda}{ \left( d+a+b \right) \left( a+b+c \right) \left( b+c+d \right) }}+{\frac {dab}{ \left( a+ b+c \right) \left( b+c+d \right) \left( a+c+d \right) }}+{\frac {abc }{ \left( b+c+d \right) \left( a+c+d \right) \left( d+a+b \right) }}\]

if easy? $x,y,z>0$ \[\frac{3}{2}\geq \sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+ \sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt { {\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\] \[\sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+\sqrt { {\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt {{\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\geq 1+4\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }}\] must be old?

In triangle,prove \[\sqrt {{\frac {ab}{ \left( b+c \right) \left( c+a \right) }}}+\sqrt { {\frac {bc}{ \left( c+a \right) \left( a+b \right) }}}+\sqrt {{\frac {ac}{ \left( a+b \right) \left( b+c \right) }}}\geq \cos \left( A \right) +\cos \left( B \right) +\cos \left( C \right) \]

xzlbq wrote: \[k=\sqrt{5}+1\] \[{\frac { \left( kz+x+y \right) \left( z+kx+y \right) \left( ky+z+x \right) }{ \left( x+y \right) \left( y+z \right) \left( z+x \right) }}\geq \frac{1}{8}\, \left( k+2 \right) ^{3}\] arqady 2005 Vietnam TST 2009

Let $a,b,c$ be positive real numbers. Prove that $$a\sqrt{\frac{b}{c+a}+\frac{1}{2}}+ b\sqrt{\frac{c}{a+b}+\frac{1}{2}}+ c\sqrt{\frac{a}{b+c}+\frac{1}{2}}\le a+b+c$$$$ab\sqrt{\frac{c}{a+b}+\frac{1}{2}}+ bc\sqrt{\frac{a}{b+c}+\frac{1}{2}}+ ca\sqrt{\frac{b}{c+a}+\frac{1}{2}}\le ab+bc+ca$$$$(a+b)\sqrt{\frac{b}{c+a}+\frac{1}{2}}+(b+c)\sqrt{\frac{c}{a+b}+\frac{1}{2}}+(c+a)\sqrt{\frac{c}{a+b}+\frac{1}{2}}\geq\frac{1}{2}( a+b+c)$$$$a\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+ b\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+ c\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\le a+b+c$$$$a^2+b^2+c^2\geq bc\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+ca\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}+ab\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\geq ab+bc+ca$$ here

Let $a,b,c$ be reals. Prove that$$(1+a^2)(1+b^2)(1+c^2)\ge(1+ ab)(1+bc )(1+ca )$$Let $a,b,c>0 $ , prove inequality$$ ( \frac{a}{b+c}+\frac{b}{c+a})(\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})\ge1.$$By C-S, $$ ( \frac{a}{b+c}+\frac{b}{c+a})\ge \frac{(a+b)^2}{2ab+bc+ca}$$$$(a+b)^2(b+c)^2(c+a)^2\ge (2ab+bc+ca) (ab+2bc+ca) (ab+bc+2ca)$$Let $ ab+bc+ca=1,$ have $$(1+a^2)(1+b^2)(1+c^2)\ge(1+ ab)(1+bc )(1+ca )$$

xzlbq wrote: if easy? $x,y,z>0$ \[\frac{3}{2}\geq \sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+ \sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt { {\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\] \[\sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+\sqrt { {\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt {{\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\geq 1+4\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }}\] must be old? \[\frac{1}{2}\,{\frac {xy+xz+2\,yz}{ \left( z+x \right) \left( x+y \right) }}\geq \sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}\] \[{\frac {xy+xz+2\,yz}{ \left( z+x \right) \left( x+y \right) }}+{ \frac {yz+xy+2\,xz}{ \left( x+y \right) \left( y+z \right) }}+{\frac {xz+yz+2\,xy}{ \left( y+z \right) \left( z+x \right) }}=3\] and \[\sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}} \left( 1+4\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }} \right) ^{-1}\geq {\frac { \left( z+2\,x+y \right) zy}{ \left( z+2\,x+y \right) zy+ \left( 2\,y+z+x \right) xz+ \left( 2\,z+x +y \right) yx}}\] Done!

sqing wrote: xzlbq wrote: https://mp.weixin.qq.com/s?__biz=MzU0NDI5MjczNw==&mid=2247484858&idx=1&sn=a12bf8f28dfa2a54b6b38b564ab136c4&chksm=fb7f2f54cc08a642f4db22b150cdb082d6fea5bb5cd88b900983dfe457e4708ae9e22b7674bf&mpshare=1&scene=1&srcid=0917Qk5xQjSlREsdsYxemVJy&pass_ticket=sCtEnHuW5lbMpMGVlgn9%2BSFmeMM3XItFo6Ls5geYFGw%3D#rd ????... 这个叙述没有问题，我说2010年我就提出了这个链接 https://artofproblemsolving.com/community/c6t243f6h380974_nicebycs_or_agxyz 这个链1#的不等式

xzlbq wrote: xzlbq wrote: if easy? $x,y,z>0$ \[\frac{3}{2}\geq \sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+ \sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt { {\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\] \[\sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+\sqrt { {\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt {{\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\geq 1+4\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }}\] must be old? \[\frac{1}{2}\,{\frac {xy+xz+2\,yz}{ \left( z+x \right) \left( x+y \right) }}\geq \sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}\] \[{\frac {xy+xz+2\,yz}{ \left( z+x \right) \left( x+y \right) }}+{ \frac {yz+xy+2\,xz}{ \left( x+y \right) \left( y+z \right) }}+{\frac {xz+yz+2\,xy}{ \left( y+z \right) \left( z+x \right) }}=3\] and \[\sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}} \left( 1+4\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }} \right) ^{-1}\geq {\frac { \left( z+2\,x+y \right) zy}{ \left( z+2\,x+y \right) zy+ \left( 2\,y+z+x \right) xz+ \left( 2\,z+x +y \right) yx}}\] Done! 非常怀疑这些结果以前也得到过，没有一个专门的书收集，总是避免不了重复，在反复重复着

xzlbq wrote: if easy? $x,y,z>0$ \[\frac{3}{2}\geq \sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+ \sqrt {{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt { {\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\] \[\sqrt {{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}}+\sqrt { {\frac {yz}{ \left( z+x \right) \left( x+y \right) }}}+\sqrt {{\frac {xz}{ \left( x+y \right) \left( y+z \right) }}}\geq 1+4\,{\frac {xyz}{ \left( y+z \right) \left( z+x \right) \left( x+y \right) }}\] must be old? 《=》 \[\frac{3}{2}\geq \sum{\sin{\frac{A}{2}}}\geq \sum{\cos{A}}\]

Nice is this: $x,y,z>0$,prove that \[{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}+{\frac {yz}{ \left( z+x \right) \left( x+y \right) }}+{\frac {xz}{ \left( x+y \right) \left( y+z \right) }}\geq \frac{1}{2}(\,{\frac {y+z}{z+2\,x+y}}+\,{ \frac {z+x}{2\,y+z+x}}+\,{\frac {x+y}{2\,z+x+y}})\] ${\frac {y+z}{z+2\,x+y}}+{\frac {z+x}{2\,y+z+x}}+{\frac {x+y}{2\,z+x+y} }$ $\geq 5/3\,{\frac {xy}{ \left( y+z \right) \left( z+x \right) }}+5/3\,{ \frac {yz}{ \left( z+x \right) \left( x+y \right) }}+5/3\,{\frac {xz} { \left( x+y \right) \left( y+z \right) }}+2\,{\frac {xyz}{ \left( y+ z \right) \left( z+x \right) \left( x+y \right) }}$

Nice is this: $x,y,z>0$ \[\sqrt {{\frac {y+z}{5\,x+2\,y+2\,z}}}+\sqrt {{\frac {z+x}{5\,y+2\,z+2 \,x}}}+\sqrt {{\frac {x+y}{5\,z+2\,x+2\,y}}}\geq \sqrt {2}\] \[{\frac {1}{59}}\,\sqrt {354}\geq \sqrt {{\frac {y+z}{77\,x+50\,y+50\,z}}}+ \sqrt {{\frac {z+x}{77\,y+50\,z+50\,x}}}+\sqrt {{\frac {x+y}{77\,z+50 \,x+50\,y}}}\]

$a,b,c,d>0$,prove \[\sqrt {{\frac {a+b+c}{7\,d+3\,a+3\,b+3\,c}}}+\sqrt {{\frac {b+c+d}{7\, a+3\,b+3\,c+3\,d}}}+\sqrt {{\frac {a+c+d}{7\,b+3\,c+3\,d+3\,a}}}+ \sqrt {{\frac {d+a+b}{7\,c+3\,d+3\,a+3\,b}}}\geq \sqrt {3}\]

$x,y,z>0$,prove \[\sqrt {{\frac { \left( y+z \right) \left( z+x \right) }{ \left( 4\,x+ y+z \right) \left( 4\,y+z+x \right) }}}+\sqrt {{\frac { \left( z+x \right) \left( x+y \right) }{ \left( 4\,y+z+x \right) \left( 4\,z+x +y \right) }}}+\sqrt {{\frac { \left( x+y \right) \left( y+z \right) }{ \left( 4\,z+x+y \right) \left( 4\,x+y+z \right) }}}\geq 1\] \[\frac{3}{2}\geq \sqrt {{\frac { \left( y+z \right) \left( z+x \right) }{ \left( z +2\,x+y \right) \left( 2\,y+z+x \right) }}}+\sqrt {{\frac { \left( z+ x \right) \left( x+y \right) }{ \left( 2\,y+z+x \right) \left( 2\,z+ x+y \right) }}}+\sqrt {{\frac { \left( x+y \right) \left( y+z \right) }{ \left( 2\,z+x+y \right) \left( z+2\,x+y \right) }}}\] \[\sqrt {{\frac { \left( z+x \right) \left( x+y \right) }{ \left( 4\,y+ z+x \right) \left( 4\,z+x+y \right) }}}\geq {\frac { \left( z+x \right) \left( x+y \right) }{ \left( z+x \right) \left( x+y \right) + \left( x+y \right) \left( y+z \right) + \left( y+z \right) \left( z +x \right) }}\]

sqing wrote: WolfusA wrote: Sqing, if you use directly AM-GM you have to prove $\frac{abc}{(a+b)(b+c)(c+a)}\ge \frac{1}{8}$ which isn't true. Could you elaborate more on your idea? By AM-GM, $$\sqrt{\frac{a}{b+c}+\frac{b}{c+a}}+\sqrt{\frac{b}{c+a}+\frac{c}{a+b}}+\sqrt{\frac{c}{a+b}+\frac{a}{b+c}}\ge 3\sqrt[6]{ (\frac{b}{c+a}+\frac{c}{a+b})( \frac{c}{a+b}+\frac{a}{b+c})( \frac{a}{b+c}+\frac{b}{c+a})}\geq3\sqrt[6]{ \frac{b+c}{\sqrt{(c+a)(a+b)}}\cdot \frac{c+a}{\sqrt{(a+b)(b+c)}}\cdot \frac{a+b}{\sqrt{(b+c)(c+a)}}}=3$$ how did you get $$3\sqrt[6]{(\frac{b}{c+a}+\frac{c}{a+b})(\frac{c}{a+b}+\frac{a}{b+c})(\frac{a}{b+c}+\frac{b}{c+a})}\geq 3\sqrt[6]{(\frac{b+c}{\sqrt{(c+a)(a+b)}})\cdot(\frac{c+a}{\sqrt{(a+b)(b+c)}})\cdot(\frac{a+b}{\sqrt{(b+c)(c+a)}})}$$??