20. Given that $a, b, x, y$ are non-negative real numbers, and $a+b=27$. Try to find the maximum value of $\lambda$ such that the inequality $\left(a x^{2}+b y^{2}+ 4 x y\right)^{3} \geqslant \lambda\left(a x^{2} y+b x y^{2}\right)^{2}$ always holds, and find the conditions for equality.
20. Let $a=0, b=27, x=27, y=2$, then the original inequality is $\lambda \leqslant 4$.

Below we prove that $\lambda=4$ makes the given inequality always true.
It is only necessary to prove under the original conditions that $\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant 4(a x+b y)^{2} x^{2} y^{2}$.
When $x$ or $y$ is 0, inequality (1) is obviously true.
When $x, y$ are both greater than 0, divide both sides of inequality (1) by $x^{3} y^{3}$ and let $t=\frac{x}{y}$, then inequality (1) becomes
$$
\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 4\left(a^{2} t+\frac{b^{2}}{t}+2 a b\right),
$$

where $t \in \mathbf{R}_{+}$. Since $\left(a t+\frac{b}{t}\right)(a+b)=a^{2} t+\frac{b^{2}}{t}+a b\left(t+\frac{1}{t}\right) \geqslant a^{2} t+\frac{b^{2}}{t}+2 a b$,
it is only necessary to prove
$$
\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 4\left(a t+\frac{b}{t}\right) \times 27 .
$$

Since $a t+\frac{b}{t}+4=a t+\frac{b}{t}+2+2 \geqslant 3 \sqrt[3]{4\left(a t+\frac{b}{t}\right)}$,
we have $\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 27 \times 4\left(a t+\frac{b}{t}\right)$. Therefore, $\lambda_{\max }=4$.
The equality in inequality (1) holds if and only if $a t+\frac{b}{t}=2, t=1$ or $x=0, b=0$ or $x=0, y=0$ or $y=0, a=0$.
Since $a+b=27>2$, the equality in inequality (1) holds if and only if $x=0, a=27, b=0$ or $y=0, a=0, b=27$ or $x=0, y=0$.