Cho hai số thực $x, y$ thỏa mãn : $9{{x}^{3}}+\left(...

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Xét hàm số , có Hay hàm đồng biến trên .
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$\Leftrightarrow \left\{ \begin{aligned}
& x\ge 0 \\
& 9{{x}^{2}}=3xy-5 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x>0; y>0 \\
& 3xy=9{{x}^{2}}+5 \\
\end{aligned} \right.$$\left( 2 \right)P={{x}^{3}}+{{y}^{3}}+6xy+3\left( 3{{x}^{2}}+1 \right)\left( x+y-2 \right)\begin{aligned}
& P={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+6xy+\left( 9{{x}^{2}}+3 \right)\left( x+y-2 \right) \\
& P={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)+6xy+\left( 3xy-2 \right)\left[ \left( x+y \right)-2 \right] \\
\end{aligned}P={{\left( x+y \right)}^{3}}-2\left( x+y \right)+4t=x+y\overset{do \left( 2 \right)}{\mathop{=}} x+\dfrac{9{{x}^{2}}+5}{3x}=4x+\dfrac{5}{3x}\ge 2\sqrt{4x.\dfrac{5}{3x}}\ge \dfrac{4\sqrt{15}}{3}.g\left( t \right)={{t}^{3}}-2t+4, t\ge \dfrac{4\sqrt{15}}{3}.g'\left( t \right)=3{{t}^{2}}-2>0,\forall t\ge \dfrac{4\sqrt{15}}{3}{{P}_{\min }}=g\left( \dfrac{4\sqrt{15}}{3} \right)=\dfrac{36+296\sqrt{15}}{9}$.