hello, use that $R=\frac{4A}{abc}$ and $a=y+z,b=x+z,c=x+y$. Sonnhard.

hello, after this you will get $2\,{\frac { \left( y+z+x \right) \left( {x}^{2}y+{x}^{2}z+x{y}^{2}-6 \,xyz+x{z}^{2}+{y}^{2}z+y{z}^{2} \right) }{ \left( y+z \right) ^{2} \left( x+z \right) ^{2} \left( x+y \right) ^{2}}} \geq 0$ which is AM-GM. Sonnhard.

Let $a, b, c$ be the sides and $R$ be the circumradius of a triangle. Prove that\[\frac{1}{4r^2} \ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\ge\frac{1}{R^2}.\] (Turkey 2005) $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le \frac{1}{4r^2}$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&p=2070790#p2070790

djb86 wrote: Let $a, b, c$ be the sides and $R$ be the circumradius of a triangle. Prove that \[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{1}{R^2}.\] Let $2s=a+b+c$ and $r$ be the inradius \[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}=\frac{2s}{4Rrs}=\frac1{2Rr} \ge \frac1{R^2}\] Last one follows from Euler's $R \ge 2r$ inequality.

We know that $R^2\geq a^2+b^2+c^2$. The inequality is equivalent with $R^2(a+b+c)\geq abc$. But we know that $(a^2+b^2+c^2)(a+b+c)\geq 9abc$.