Question

[My Phạm]

CM: \(\frac{x^6+y^9}{4}\)>=3x²y³-16 vs x,y >=0

Answer 1

[Thiên An]

Vì x, y > =0 theo BĐT Cô-si

\(x^6+y^9=\frac{1}{4}x^6+\frac{1}{4}x^6+\frac{1}{4}x^6+\frac{1}{4}x^6+\frac{1}{4}y^9+\frac{1}{4}y^9+\frac{1}{4}y^9+\frac{1}{4}y^9+16+16+16+16-64\)

\(\ge12\sqrt[12]{\left(\frac{1}{4}x^6\right)^4.\left(\frac{1}{4}y^9\right)^4.16^4}-64=12\sqrt[12]{x^{24}y^{36}}-64=12x^2y^3-64\)

\(\Rightarrow\frac{x^6+y^9}{4}\ge\frac{12x^2y^3-64}{4}=3x^2y^3-16\)

Đẳng thức xảy ra  \(\Leftrightarrow\)  \(\frac{1}{4}x^6=\frac{1}{4}y^9=16\)  \(\Leftrightarrow\)  \(\hept{\begin{cases}x=2\\y=\sqrt[9]{64}\end{cases}}\)