Do \(p=\dfrac{a+b+c}{2}\Rightarrow2p=a+b+c\)
Ta có: \(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{a+b+c-\left(a+b\right)}=\dfrac{4}{c}\)
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{2p-\left(b+c\right)}=\dfrac{4}{a}\)
\(\dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{4}{2p-\left(a+c\right)}=\dfrac{4}{b}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Dấu "=" xảy ra khi \(a=b=c\) hay tam giác là tam giác đều
Ta có : \(p=\frac{a+b+c}{2}\Rightarrow2p=a+b+c\)
Do a ; b ; c là 3 cạnh tam giác \(\Rightarrow b+c-a;c+a-b;a+b-c>0\)
\(b+c-a>0\Rightarrow\frac{b+c}{2}-\frac{a}{2}>0\Rightarrow\frac{a+b+c}{2}-a>0\Rightarrow p-a>0\)
CMTT , ta có : \(p-b>0;p-c>0\)
Áp dụng BĐT phụ \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) với x ; y > 0 vào bài toán , ta có
\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{a+b+c-a-b}=\frac{4}{c}\left(1\right)\)
CMTT : \(\frac{1}{p-a}+\frac{1}{p-c}\ge\frac{4}{b};\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) \(\left(đpcm\right)\)
