bđt<=>\(S_a\left(a-b\right)^2+S_b\left(b-c\right)^2+S_c\left(c-a\right)^2\ge0\)
with \(S_a=\frac{1}{2\left(a^2+b^2\right)}-\frac{c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(S_b=\frac{1}{2\left(b^2+c^2\right)}-\frac{a}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(S_c=\frac{1}{2\left(c^2+a^2\right)}-\frac{b}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
cần cm \(S_a+S_c;S_b+S_c>0\)
lại có:\(S_a+S_c=\frac{1}{2}\left(\frac{1}{a^2+b^2}+\frac{1}{c^2+a^2}\right)-\frac{1}{\left(a+b\right)\left(c+a\right)}\)
\(>\frac{1}{2}\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(c+a\right)^2}\right)-\frac{1}{\left(a+b\right)\left(c+a\right)}>0\)
cmtt=>q.e.d