Gọi p là nửa chu vi tam giác đó \(\Rightarrow p=\frac{a+b+c}{2}\)
Ta có : \(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{a+c-b}=\frac{2}{p-a}+\frac{2}{p-b}+\frac{2}{p-c}\)
Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)được :
\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\)
Tương tự : \(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\) ; \(\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{b}\)
Cộng các bất đẳng thức trên theo vế : \(2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Leftrightarrow2\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
