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Question

Cho x>1, y>0, chứng minh :

$\frac{1}{\left(x-1\right)^3}+\left(\frac{x-1}{y}\right)^3+\frac{1}{y^3}\ge3\left(\frac{3-2x}{x-1}+\frac{x}{y}\right)$

Hung nguyen · Accepted Answer

Ta có:

$\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1$

$\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)$

$\Rightarrow\dfrac{1}{\left(x-1\right)^3}+\left(\dfrac{x-1}{y}\right)^3+\dfrac{1}{y^3}\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)$

$=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)$Câu trả lời: Ta có:

$\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1$

$\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)$

$\Rightarrow\dfrac{1}{\left(x-1\right)^3}+\left(\dfrac{x-1}{y}\right)^3+\dfrac{1}{y^3}\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)$

$=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)$