Question

Cho x>1,y>1.
 CM: P= \(\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\cdot\left(x-1\right).\left(y-1\right)}\ge8.\)

Answer 1

Ta có \(P=\frac{\left(x^3+y^3\right)-\left(x^2+y^2\right)}{\left(x-1\right)\left(y-1\right)}=\frac{x^2\left(x-1\right)+y^2\left(y-1\right)}{\left(x-1\right)\left(y-1\right)}=\frac{x^2}{y-1}+\frac{y^2}{x-1}\)

Áp dụng bđt AM-GM ta có \(\frac{x^2}{y-1}+\frac{y^2}{x-1}\ge\frac{\left(x+y\right)^2}{x+y-2}\)

Đặt \(t=x+y\)

Xét \(\frac{t^2}{t-2}\ge8\Leftrightarrow t^2\ge8t-16\Leftrightarrow t^2-8t+16\ge0\Leftrightarrow\left(t-4\right)^2\ge0\)luôn đúng

Vậy \(\frac{\left(x+y\right)^2}{x+y-2}\ge8\) hay \(P\ge8\).