Question

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

\(\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}\ge a+b+c\\ \)

Answer 1

Đặt a+b-c=x

       b+c-a=y

      c+a-b=z

\(A=\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}\)

Ta có a;b;c là độ dài 3 cạnh tam giác nên x;y;z>0

\(4A=\frac{2a.2b}{x}+\frac{2b.2c}{y}+\frac{2c.2a}{z}\)

\(=\frac{\left(x+z\right)\left(x+y\right)}{x}+\frac{\left(x+y\right)\left(y+z\right)}{y}+\frac{\left(x+z\right)\left(y+z\right)}{z}\)

\(=3\left(x+y+z\right)+\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\right)\)

\(\ge3\left(x+y+z\right)+\frac{\left(x+y+z\right)xyz}{xyz}\)\(=4\left(x+y+z\right)=4\left(a+b+c\right)\)  (Do x;y;z>0)

\(\Rightarrow A\ge a+b+c\)