Estonian Mathematical Olympiad

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.

---

Alternative solution.

Let $f(x)=x^3+8x^2-19x+10$. The given inequality is equivalent to $f(x)\ge 0$. Note that $f^{\prime }(x)=3x^2+16x-19$. As $D=16^2-4\cdot 3\cdot (-19)>0$, the function $f^{\prime }$ 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 $f^{\prime }(x)$ is positive, $f^{\prime }$ is increasing at 1 whence the values of $f^{\prime }$ 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

---

Alternative solution.

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.