$\sum_{\text{cyc}}\sqrt{\frac{a}{-a+b+c}} \ge 3\sqrt[6]{\frac{abc}{(b+c-a)(c+a-b)(a+b-c)}}\ge 3$

Can someone explain me the last step??

It's Schur: $$abc\geq\prod_{cyc}(a+b-c)$$it's $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0.$$

AlastorMoody wrote: Can someone explain me the last step?? https://artofproblemsolving.com/community/c6h1300133p6921601

CatalinBordea wrote: If $ a,b,c $ represent the lengths of the sides of a triangle, prove the inequality: $$ 3\le\sum_{\text{cyc}}\sqrt{\frac{a}{-a+b+c}} . $$ stronger is: $$ 3\le\sum_{\text{cyc}}\sqrt{\frac{a}{-a+b+c}}\cos{\frac{B-C}{2}} . $$ easy prove \[\sqrt {{\frac {a}{b+c-a}}}\cos{\frac{B-C}{2}}\geq 3\,{\frac { \left( b+c \right) \left( a+b- c \right) \left( a+c-b \right) }{ \left( b+c \right) \left( a+b-c \right) \left( a+c-b \right) + \left( c+a \right) \left( b+c-a \right) \left( a+b-c \right) + \left( a+b \right) \left( a+c-b \right) \left( b+c-a \right) }}\]

you can do this: \[2 \left( \sqrt {{\frac {R}{r}}}-\sqrt {2} \right)\geq \sqrt {{\frac {a}{b+c-a}}}+\sqrt {{\frac {b}{a+c-b}}}+\sqrt {{\frac {c }{a+b-c}}}-3\geq \sqrt {2} \left( \sqrt {{\frac {R}{r}}}-\sqrt {2} \right)\]

$a=x+y,b=y+z,c=z+x$ $\sum{\sqrt{\frac{x+y}{z}}} \geq 3\sqrt[6]{\frac{2xyz+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{xyz}}=3\sqrt[6]{2+\sum{\frac{x}{y}}} \geq 3\sqrt[6]{8}=3\sqrt2$