Problem 12.3. Find all real numbers $a$, such that the inequality

$$
x^{4}+2 a x^{3}+a^{2} x^{2}-4 x+3>0
$$

holds true for all real numbers $x$.

Nikolai Nikolov
12.3. First Solution. Write the equation in the form

$$
x^{2}(x+a)^{2}>4 x-3
$$

Then for $x=1$ we get $(a+1)^{2}>1$, i.e. $a>0$ or $a-4 a-3$. Thus, $a>0$.

Conversely, if $a>0$, then (1) is satisfied for all $x$. Indeed, when $x \leq 0$ this is obvious and when $x>0$ we have $x^{2}(x+a)^{2}>x^{4} \geq 4 x+3$, since the later inequality is equivalent to $(x-1)^{2}\left((x+1)^{2}+2\right) \geq 0$.

Second Solution. It follows from (1) that we have to find all a, for which

$$
a\frac{\sqrt{4 x-3}}{x}-x=g(x)
$$

for all $x \geq \frac{4}{3}$.

The first case is impossible since $\lim _{x \rightarrow+\infty} f(x)=-\infty$. The maximum of the function $g$ equals 0 (and it is attained for $x=1$ ). Therefore the answer is $a>0$.