Dễ dàng chứng minh bất đẳng thức phụ :
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a;b>0\)và p - a; p - b; p - c > 0 theo bất đẳng thức trong tam giác.
Áp dụng bất đẳng thức phụ vừa chứng minh, ta có:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-a-b}=\dfrac{4}{c}\left(1\right)\)
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{2p-b-c}=\dfrac{4}{a}\left(2\right)\)
\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{2p-c-a}=\dfrac{4}{a}\left(3\right)\)
Cộng (1); (2); (3) theo vế, ta có:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\RightarrowĐPCM\)
Ta CM BĐT sau :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy ; ta có :
\(\left(x-y\right)^2\ge0\\ \Rightarrow x^2-2xy+y^2\ge0\\ \Rightarrow x^2+y^2\ge2xy\\ \Rightarrow\left(x+y\right)^2\ge4xy\\ \Rightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\\ \Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\left(đpcm\right)\)
\(\Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{c}\\ \dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\\ \dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{4}{b}\\ \Rightarrow2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\\ \Rightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(đpcm\right)\)
