Cho a,b,c là 3 cạnh của một tam giác. Chứng minh rằng [tex]\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}\geq(a+b+c)[/tex].
đặt [tex]a+b-c=x\\\\ b+c-a=y\\\\ c+a-b=z\\\\ => x+z=2a; x+y=2b\\\\ => 4ab=(x+z).(x+y)\\\\ => ab=\frac{(x+z).(x+y)}{4} => \frac{ab}{a+b-c}=\frac{(x+z).(x+y)}{4x}\\\\ =\frac{x^2+xz+yz+xy}{4x}=\frac{x}{4}+\frac{z}{4}+\frac{yz}{4x}+\frac{y}{4}\\\\ =(\frac{x}{4}+\frac{z}{4}+\frac{y}{4})+\frac{yz}{4x}[/tex]
CMTT => [tex]\frac{bc}{b+c-a}=(\frac{x}{4}+\frac{z}{4}+\frac{y}{4})+\frac{xz}{4y}\\\\ \frac{ca}{c+a-b}=(\frac{x}{4}+\frac{z}{4}+\frac{y}{4})+\frac{xy}{4z}\\\\ => \frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ca}{c+a-b}=3.(\frac{x}{4}+\frac{z}{4}+\frac{y}{4})+\frac{1}{4}.(\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z})[/tex]
giờ cần CM: [tex]\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}\geq x+y+z\\\\[/tex]
áp dụng BĐT Cô-si cho các số dương, có:
[tex]\frac{yz}{x}+\frac{xz}{y}\geq 2\sqrt{\frac{yz}{x}.\frac{xz}{y}}=2z\\\\ \frac{xz}{y}+\frac{xy}{z}\geq 2\sqrt{\frac{xz}{y}.\frac{xy}{z}}=2x\\\\ \frac{yz}{x}+\frac{xy}{z}\geq 2\sqrt{\frac{yz}{x}.\frac{xy}{z}}=2y[/tex]
cộng từng vế BĐT ta có đpcm
dấu "=" xảy ra <=> x=y=z <=> a=b=c