Proof We denote by $x_1, x_2, x_3$ the three positive roots of the polynomial $\varphi(x) = ax^3 + bx^2 + cx + d$. By Vieta's theorem, we have
$$\begin{align*} x_1 + x_2 + x_3 &= -\frac{b}{a}, \\ x_1x_2 + x_2x_3 + x_3x_1 &= \frac{c}{a}, \\ x_1x_2x_3 &= -\frac{d}{a}. \end{align*}$$
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As $\varphi(0) < 0$, we get $d < 0$, and hence $a > 0$.
Dividing both sides of ① by $a^3$, we get an equivalent form of ①:
$$\begin{aligned} & 7\left(-\frac{b}{a}\right)\frac{c}{a} \le 2\left(-\frac{b}{a}\right)^3 + 9\left(-\frac{d}{a}\right) \\ \Leftrightarrow & 7(x_1 + x_2 + x_3)(x_1x_2 + x_2x_3 + x_3x_1) \\ & \le 2(x_1 + x_2 + x_3)^3 + 9x_1x_2x_3 \\ \Leftrightarrow & x_1^2x_2 + x_1^2x_3 + x_2^2x_1 + x_2^2x_3 + x_3^2x_1 + x_3^2x_2 \\ & \le 2(x_1^3 + x_2^3 + x_3^3). \qquad \textcircled{2} \end{aligned}$$
Because $x_1, x_2, x_3$ are all greater than 0, $(x_1 - x_2)(x_1^2 - x_2^2) \ge 0$. That is to say,
$$x_1^2 x_2 + x_2^2 x_1 \le x_1^3 + x_2^3.$$
By the same argument, $x_2^2 x_3 + x_3^2 x_2 \le x_2^3 + x_3^3$, $x_3^2 x_1 + x_1^2 x_3 \le x_3^3 + x_1^3$.
Summing up these three inequalities we get ②, and the
equality holds iff $x_1 = x_2 = x_3$.