[tex]\dfrac1{a+b}+\dfrac1{b+c}+\dfrac1{c+a}>\dfrac3{a+b+c}\qquad\ \ /\cdot(a+b+c) \\\\ \dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}>3\\\\ \dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b+c}{b+c}+\dfrac{b}{c+a}+\dfrac{a+c}{c+a}>3 \\\\ 1+\dfrac{c}{a+b}+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1>3\qquad\ \ /-3 \\\\ \dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}>0[/tex]
[tex]\left \big {a>0}\\ \atop \left \big{b>0}\\\atop\big{ c>0}\right\right\}\implies\, \left \big {\frac c{a+b}>0}\\ \atop \left \big{\frac a{b+c}>0}\\ \atop\big{\frac b{c+a}>0}\right\right\} \implies\ \ \dfrac c{a+b}+\dfrac a{b+c}+\dfrac b{c+a}>0}\right\right\}[/tex]
co należało wykazać.