Question

Cho a,b,c là độ dài 3 cạnh một tam giác. Tìm:

\(MinP=\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}\)

Answer 1

Đặt \(b+c-a=2x;c+a-b=2y;a+b-c=2z\) \(\Rightarrow a=y+z;b=x+z;c=x+y\)

\(\dfrac{4a}{b+c-a}+\dfrac{4b}{c+a-b}+\dfrac{4c}{a+b-c}=\dfrac{4\left(y+z\right)}{2x}+\dfrac{4\left(x+z\right)}{2y}+\dfrac{4\left(x+y\right)}{2z}\)\(=\dfrac{2\left(y+z\right)}{x}+\dfrac{2\left(x+z\right)}{y}+\dfrac{2\left(x+y\right)}{z}=2\left(\dfrac{y}{x}+\dfrac{z}{x}+\dfrac{x}{y}+\dfrac{z}{y}+\dfrac{x}{z}+\dfrac{y}{z}\right)\ge2.\left(2+2+2\right)=12\)