Solve 5(f)+b+c=0,text{thenprovethat}(b+c)^2/3bc+(c+a)^2/3ca+(a+b)^2/3ab=1

$$5 ( f ) + b + c = 0 , \text { then prove that } \frac { ( b + c ) ^ { 2 } } { 3 b c } + \frac { ( c + a ) ^ { 2 } } { 3 c a } + \frac { ( a + b ) ^ { 2 } } { 3 a b } = 1$$

Solve for f, b, c, a

$$f=\frac{-b-c}{5}\text{, }b\in \mathrm{R}\setminus -c,0\text{, }c\neq 0\text{, }a=-\frac{bc}{b+c}$$

$$f=\frac{a}{5}\text{, }b=-\left(a+c\right)\text{, }c\in \mathrm{R}\setminus 0,-a\text{, }a\neq 0$$