Question

Cho ba số dương a,b,c sao cho : abc=1 .Chứng minh :

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{3}{2}\left(a+b+c-1\right)\)

Answer 1

Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)

BĐT trở thành:

\(\dfrac{y^2}{xz}+\dfrac{z^2}{xy}+\dfrac{x^2}{yz}\ge\dfrac{3}{2}\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}-1\right)\)

\(\Leftrightarrow2\left(x^3+y^3+z^3\right)+3xyz\ge3x^2y+3y^2z+3z^2x\)

Áp dụng BĐT Schur ta có:

\(x^3+y^3+z^3+3xyz\ge x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\)

\(\Rightarrow VT\ge\left(x^3+xy^2\right)+\left(y^3+yz^2\right)+\left(z^3+zx^2\right)+x^2y+y^2z+z^2x\ge3\left(x^2y+y^2z+z^2x\right)\)