Cho a, b, c là độ dài 3 cạnh tam giác. Chứng minh rằng:

$\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}\ge a+b+c$.

Đặt:  $\left\{\begin{array}{l}x=a+b-c\\ y=b+c-a\\ z=c+a-b\end{array}\right.$

$\Rightarrow \left\{\begin{array}{l}x+y=\left(a+b-c\right)+\left(b+c-a\right)=2b\\ y+z=\left(b+c-a\right)+\left(c+a-b\right)=2c\\ x+z=\left(a+b-c\right)+\left(c+a-b\right)=2a\end{array}\right.$

Với a, b, c là ba cạnh của tam giác, thì:

$\left\{\begin{array}{l}b+c>a\\ a+c>b\\ a+b>c\end{array}\right.\Rightarrow \left\{\begin{array}{l}b+c-a>0\\ a+c-b>0\\ a+b-c>0\end{array}\right.\Rightarrow \left\{\begin{array}{l}y>0\\ z>0\\ x>0\end{array}\right.$

Khi đó  $A=\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}$

$\Rightarrow 4A=\frac{2a.2b}{a+b-c}+\frac{2b.2c}{b+c-a}+\frac{2c.2a}{c+a-b}$

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

$=\frac{x^2+x\left(y+z\right)+yz}{x}+\frac{y^2+y\left(z+x\right)+zx}{y}+\frac{z^2+z\left(x+y\right)+xy}{z}$

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

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

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

Áp dụng BĐT Cô-si ta có:

$\frac{{\left(yz\right)}^2}{xyz}+\frac{{\left(zx\right)}^2}{xyz}\ge 2\frac{\sqrt{{\left(yz\right)}^2.{\left(zx\right)}^2}}{xyz}=2\frac{z.xyz}{xyz}=2z$

$\frac{{\left(zx\right)}^2}{xyz}+\frac{{\left(xy\right)}^2}{xyz}\ge 2\frac{\sqrt{{\left(zx\right)}^2.{\left(xy\right)}^2}}{xyz}=2\frac{x.xyz}{xyz}=2x$

$\frac{{\left(xy\right)}^2}{xyz}+\frac{{\left(yz\right)}^2}{xyz}\ge 2\frac{\sqrt{{\left(xy\right)}^2.{\left(yz\right)}^2}}{xyz}=2\frac{y.xyz}{xyz}=2y$

Suy ra  $\frac{{\left(yz\right)}^2}{xyz}+\frac{{\left(zx\right)}^2}{xyz}+\frac{{\left(xy\right)}^2}{xyz}\ge \frac{2z+2x+2y}{2}=x+y+z$

Khi đó:

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

Þ 4A ≥ 4(x + y + z)

Þ A ≥ (x + y + z) = a + b + c

Dấu “=” xảy ra khi và chỉ khi: a = b = c.

Vậy  $\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}\ge a+b+c$ khi và chỉ khi a = b = c.