Question

cho a,b,c>1
cm: a^2/b-1+b^2/c-1+c^2/a-1>=12

Answer 1

Đặt \(\left(a-1;b-1;c-1\right)=\left(x;y;z\right)\Rightarrow x;y;z>0\)

\(P=\dfrac{\left(x+1\right)^2}{y}+\dfrac{\left(y+1\right)^2}{z}+\dfrac{\left(z+1\right)^2}{x}\ge\dfrac{\left(x+y+z+3\right)^2}{x+y+z}\)

\(P\ge\dfrac{\left(x+y+z\right)^2+6\left(x+y+z\right)+9}{x+y+z}=x+y+z+\dfrac{9}{x+y+z}+6\ge2\sqrt{\dfrac{9\left(x+y+z\right)}{x+y+z}}+6=12\)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=2\)