Do p là nửa chu vi của tam giác
\(\Rightarrow p=\frac{a+b+c}{2}\)
Xét: \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\)
\(\Leftrightarrow\frac{1}{\frac{a+b+c}{2}-a}+\frac{1}{\frac{a+b+c}{2}-b}+\frac{1}{\frac{a+b+c}{2}-c}\)
\(\Leftrightarrow\frac{1}{\frac{b+c-a}{2}}+\frac{1}{\frac{a+c-b}{2}}+\frac{1}{\frac{a+b-c}{2}}\)
\(\Leftrightarrow\frac{2}{b+c-a}+\frac{2}{a+c-b}+\frac{2}{a+b-c}\)
\(\Leftrightarrow2\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\)
Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) với a , b > 0
\(\Rightarrow\left\{\begin{matrix}\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{2}{c}\\\frac{1}{a+c-b}+\frac{1}{a+b-c}\ge\frac{2}{a}\\\frac{1}{b+c-a}+\frac{1}{a+b-c}\ge\frac{2}{b}\end{matrix}\right.\)
Cộng theo từng vế:
\(\Rightarrow2\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Leftrightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) ( đpcm )
