BĐT cô si: \(\dfrac{x+y}{2}>\left(hoặc=\right)\sqrt{xy}\)
=>x+y >(hoặc =) \(2\sqrt{xy}\)
=>\(\left(x+y\right)^2>\left(hoặc=\right)4xy\)
=>\(\dfrac{1}{x}+\dfrac{1}{y}>\left(hoặc=\right)\dfrac{4}{x+y}\)
vì P=\(\dfrac{a+b+c}{2}=>a+b+c=2p\)
=>c=2p-a-b
b=2p-a-c
a=2p-b-c
ta có:\(\dfrac{1}{p-a}+\dfrac{1}{p-b}>hoặc=\dfrac{4}{p-a+p-b}=\dfrac{4}{c}\)
\(\dfrac{1}{p-a}+\dfrac{1}{p-c}>\left(hoặc=\right)\dfrac{4}{p-a+p-c}=\dfrac{4}{b}\)
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}>\left(hoặc=\right)\dfrac{4}{p-b+p-c}=\dfrac{4}{a}\)
cộng vế với vế ta đc
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)>\left(hoặc=\right)4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
<=>\(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}>\left(hoặc=\right)2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
