Given $a,b,c,d$ are positive real number satisfy $abc+bcd+cad+bad=1$. Find the value of minimize $$P=4(a^3+b^3+c^3)+9d^3$$
By AM-GM: $a^3+b^3+c^3 \geq 3abc~~(1)$ $k^3d^3+a^3+b^3 \geq 3kabd$ $\Rightarrow k^2d^3+a^3.\dfrac{1}{k}+b^3.d\frac{1}{k} \geq 3abd~~(2)$
Similar: $k^2d^3+a^3.\dfrac{1}{k}+c^3.\dfrac{1}{k} \geq 3acd~~(3)$
$k^2d^3+b^3.\dfrac{1}{k}+c^3.\dfrac{1}{k} \geq 3bcd~~(4)$
$3k^2d^3+(a^3+b^3+c^3)(1+\dfrac{1}{k}+\dfrac{1}{k}) \geq 3(abc+abd+acd+bcd)=3$ $$\Rightarrow 3k^2d^3+(a^3+b^3+c^3).\dfrac{k+2}{k} \geq 3$$
Can find $k$ is positive number satisfy $$\dfrac{3k^2}{\dfrac{k+2}{k}}=\dfrac{9}{4}$$
I need new method
Let $m>0$ and $n>0$.
Hence, by AM-GM $$\sum_{cyc}(ma^3+mb^3+3d^3)\geq3\sqrt[3]{3m^2}(abd+acd+bcd)$$ and $$n(a^3+b^3+c^3)\geq3nabc.$$ Thus, we have a system:
$3\sqrt[3]{3m^2}=3n$ and $2m+n=4$.
Since the equality occurs for $a=b=c$ and $ma^3=3d^3$, the rest is smooth.