Question

[Lê Vũ Anh Thư]

Cho a, b, c là 3 cạnh của tam giác. CMR: \(\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ac}{a+c-b}\ge a+b+c\)

Answer 1

[Nguyễn Hưng Phát]

hmm..
Đặt \(\left(x;y;z\right)=\left(a+b-c;b+c-a;c+a-b\right)\)

\(\Leftrightarrow\hept{\begin{cases}a=\frac{x+z}{2}\\b=\frac{x+y}{2}\\c=\frac{y+z}{2}\end{cases}}\)

Bất đẳng thức cần chứng minh tương đương với:

\(\frac{\left(x+y\right)\left(x+z\right)}{4x}+\frac{\left(y+z\right)\left(x+y\right)}{4y}+\frac{\left(z+x\right)\left(z+y\right)}{4z}\ge x+y+z\)

Ta có:\(\frac{\left(x+y\right)\left(x+z\right)}{4x}+\frac{\left(y+z\right)\left(x+y\right)}{4y}+\frac{\left(z+x\right)\left(z+y\right)}{4z}\)

\(=\frac{x^2+xy+xz+yz}{4x}+\frac{xy+yz+y^2+zx}{4y}+\frac{zx+zy+z^2+xy}{4z}\)

\(=\frac{3\left(x+y+z\right)}{4}+\frac{1}{4}\left(\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\right)\)\(=\frac{3}{4}\left(x+y+z\right)+\frac{1}{4}\left(\frac{y^2z^2}{xyz}+\frac{z^2x^2}{xyz}+\frac{x^2y^2}{xyz}\right)\)

\(\ge\frac{3}{4}\left(x+y+z\right)+\frac{1}{4}\left[\frac{\left(xy+yz+zx\right)^2}{3xyz}\right]\)\(\ge\frac{3}{4}\left(x+y+z\right)+\frac{1}{4}\left[\frac{3xyz\left(x+y+z\right)}{3xyz}\right]\)

\(=x+y+z\)

Bất đẳng thức đã được chứng minh.

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)