Dễ dàng c/m : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\forall x;y>0\) (*)
a ; b ; c là độ dài 3 cạnh của \(\Delta\Rightarrow\left\{{}\begin{matrix}a;b;c>0\\a+b-c;b+c-a;c+a-b>0\end{matrix}\right.\)
AD (*) ta được : \(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{2b}=\dfrac{2}{b}\)
CMTT : \(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c};\dfrac{1}{c+a-b}+\dfrac{1}{a+b-c}\ge\dfrac{2}{a}\)
Suy ra : \(2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
Do a;b;c là độ dài 3 cạnh của 1 tam giác \(\Rightarrow\left\{{}\begin{matrix}a+b-c>0\\b+c-a>0\\c+a-b>0\end{matrix}\right.\)
Ta có:
\(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}=\dfrac{2}{b}\)
Tương tự:
\(\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{2}{a}\) ; \(\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{2}{c}\)
Cộng vế:
\(\dfrac{2}{a+b-c}+\dfrac{2}{b+c-a}+\dfrac{2}{c+a-b}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\)
\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu "=" xảy ra khi \(a=b=c\)
