Đặt: \(b+c-a=x\)
\(a+c-b=y\)
\(a+b-c=z\)
Suy ra:
\(2a=y+z\)
\(2b=x+z\)
\(2c=x+y\)
Ta có:
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(=\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\) ( BĐT luôn đúng)
=> ĐPCM
a,b,c là độ dài 3 cạnh t/g
\(\Rightarrow\dfrac{a}{b+c-a};\dfrac{b}{a+c-b};\dfrac{c}{a+b-c}>0\)
\(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(A+\dfrac{3}{2}=\dfrac{a}{b+c-a}+\dfrac{1}{2}+\dfrac{b}{a+c-b}+\dfrac{1}{2}+\dfrac{c}{b+a-c}+\dfrac{1}{2}\)
\(A+\dfrac{3}{2}=\dfrac{a+b+c}{2\left(b+c-a\right)}+\dfrac{a+b+c}{2\left(a+c-b\right)}+\dfrac{a+b+c}{2\left(b+a-c\right)}\)
\(A+\dfrac{3}{2}=\dfrac{\left(a+b+c\right)}{2}\left(\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}+\dfrac{1}{b+a-c}\right)\)
\(A+\dfrac{3}{2}\ge\dfrac{a+b+c}{2}\cdot\dfrac{9}{b+c-a+a+c-b+b+a-c}\)
\(A+\dfrac{3}{2}\ge\dfrac{9}{2}\)
\(\Rightarrow A\ge3\left(đpcm\right)\)
