The given inequality is equivalent to $x^3 + 8x^2 - 19x + 10 \ge 0$. Note that $x^3 + 8x^2 - 19x + 10 = (x-1)^2(x+10)$. As $(x-1)^2 \ge 0$ and $x+10 > 0$ for positive $x$, this inequality holds indeed.

Let $f(x) = x^3 + 8x^2 - 19x + 10$. The given inequality is equivalent to $f(x) \ge 0$. Note that $f'(x) = 3x^2 + 16x - 19$. As $D = 16^2 - 4 \cdot 3 \cdot (-19) > 0$, the function $f'$ has two real roots. Let the smaller and the larger root be $x_1$ and $x_2$, respectively; then $x_2 = 1$ and $x_1 < 0$. As the coefficient of the quadratic term of $f'(x)$ is positive, $f'$ is increasing at 1 whence the values of $f'$ are negative in the interval $(x_1; 1)$ and positive in the interval $(1; \infty)$. Hence $f$ is decreasing in the interval $(x_1; 1)$ and increasing in the interval $(1; \infty)$ (Fig. 39). Thus $f(x) \ge f(1) = 0$ holds for all $x$ in the interval $(x_1; \infty)$ that includes every positive $x$.

Fig. 39

For every positive integer $k$, the AM-GM inequality implies $\frac{x+1 \cdot k}{k+1} \ge \sqrt[k+1]{x \cdot 1^k}$. Using this inequality for $k=1, k=2$, and $k=5$, we obtain the inequalities $x+1 \ge 2\sqrt{x}$, $x+2 \ge 3\sqrt[3]{x}$, $x+5 \ge 6\sqrt[6]{x}$, respectively. The desired result is now obtained by multiplying the corresponding sides.