
Problem 1. Find the values of the real parameter $a$, for which the inequality $x^{6}-6 x^{5}+$ $12 x^{4}+a x^{3}+12 x^{2}-6 x+1 \geq 0$ is satisfied for all real $x$.


Solution. When $x=0$ the given inequality is satisfied for all $a$. Thus, it is enough to find such $a$, that $\left(x^{3}+\frac{1}{x^{3}}\right)-6\left(x^{2}+\frac{1}{x^{2}}\right)+12\left(x+\frac{1}{x}\right)+a \geq 0$ for all $x>0$ and $\left(x^{3}+\frac{1}{x^{3}}\right)-$ $6\left(x^{2}+\frac{1}{x^{2}}\right)+12\left(x+\frac{1}{x}\right)+a \leq 0$ for all $x0 \Longleftrightarrow t \geq 2$ and $x<0 \Leftrightarrow t \leq-2$. But $x^{2}+\frac{1}{x^{2}}=t^{2}-2$ and $x^{3}+\frac{1}{x^{3}}=t^{3}-3 t$.

We consider the function $f(t)=\left(t^{3}-3 t\right)-6\left(t^{2}-2\right)+12 t+=t^{3}-6 t^{2}+9 t+12+a$. The problem is reduced to find such $a$, that $f(t) \geq 0$ for all $t \geq 2$ and $f(t) \leq 0$ for all $t \leq-2$ simultaneously.

Figure 14.

![](https://cdn.mathpix.com/cropped/2024_06_03_e679746950d7d3832835g-143.jpg?height=151&width=377&top_left_y=762&top_left_x=446)

But $f^{\prime}(t)=3 t^{2}-12 t+9=3(t-1)(t-3)$, from where it is easy to obtain that (Figure 14) $f(t) \geq 0$ for all $t \geq 2$ iff $f(3) \geq 0$ and $f(t) \leq 0$ for all $t \leq-2$ iff $f(-2) \leq 0$. Since $f(3)=a+12$ and $f(-2)=a-38$, we find for $a: a \geq-12$ and $a \leq 38$. Finally, $-12 \leq a \leq 38$.
