CM theo chiều ngược lại , nếu a ; b ; c là 3 cạnh tam giác
thì tổng các phân thức trên > 1 ( 1 )
\(\frac{a^2+b^2-c^2}{2ab}+1=\frac{\left(a+b\right)^2-c^2}{2ab}\) ; \(\frac{b^2+c^2-a^2}{2bc}-1=\frac{\left(b-c\right)^2-a^2}{2bc}\) ;
\(\frac{c^2+a^2-b^2}{2ac}-1=\frac{\left(c-a\right)^2-b^2}{2ac}\)
\(\Rightarrow\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}-1=\frac{\left(a+b\right)^2-c^2}{2ab}+\frac{\left(b-c\right)^2-a^2}{2bc}+\frac{\left(c-a\right)^2-b^2}{2ac}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{2ab}+\frac{\left(b-c-a\right)\left(b-c+a\right)}{2bc}+\frac{\left(c-a-b\right)\left(c-a+b\right)}{2ac}\)
\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{2ab}+\frac{\left(b-c-a\right)\left(b-c+a\right)}{2bc}+\frac{\left(a+b-c\right)\left(a-c-b\right)}{2ac}\)
\(=\left(a+b-c\right)\left(\frac{a+b+c}{2ab}+\frac{b-c-a}{2bc}+\frac{a-c-b}{2ac}\right)\)
\(=\left(a+b-c\right)\left[\frac{\left(a+b+c\right)c+\left(b-c-a\right)a+\left(a-c-b\right)b}{2abc}\right]\)
\(=\left(a+b-c\right)\left[\frac{ac+bc+c^2+ab-ac-a^2+ab-bc-b^2}{2abc}\right]\)
\(=\left(a+b-c\right)\left[\frac{c^2-\left(a-b\right)^2}{2abc}\right]\)
\(=\left(a+b-c\right).\frac{\left(c-a+b\right)\left(c+a-b\right)}{2abc}\) ( * )
Vì a ; b ; c là 3 cạnh của tam giác nên biểu thức (*) luôn > 0
\(\Rightarrow\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}-1>0\)
\(\Rightarrow\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}>1\left(đpcm\right)\) ( 2 )
Từ ( 1 ) ; ( 2 ) => a ; b ; c là 3 cạnh của 1 tam giác
