China Southeastern Mathematical Olympiad

Denote $f(x)=\frac{a{x}^2+b}{\sqrt{x^2+1}}$. It is easy to see that $a>0$. By $f(0)=b$, we see that $b\ge 3$.

a. If $b-2a\ge 0$, $$f(x)=\frac{a{x}^2+b}{\sqrt{x^2+1}}=a\sqrt{x^2+1}+\frac{b-a}{\sqrt{x^2+1}}\ge 2\sqrt{a(b-a)}=3,$$ equality holds if $a\sqrt{x^2+1}=\frac{b-a}{\sqrt{x^2+1}}$, that is, if $x=\pm \sqrt{\frac{b-2a}{a}}$. The value of $a$ for given $b$ is $a=\frac{b-\sqrt{b^2-9}}{2}$, especially when $b=3$, $a=\frac{3}{2}$.

b. If $b-2a<0$, let $\sqrt{x^2+1}=t$ ($t\ge 1$). $f(x)=g(t)=at+\frac{b-a}{t}$ is monotonically increasing when $t\ge 1$, so, $$\underset{x\in \mathbb{R}}{\min }f(x)=g(1)=a+b-a=b=3,\text{ when }a>\frac{3}{2}.$$

Summing up, we get (1) the range of $b$ is $[3,+\infty )$. (2) If $b=3$, then $a\ge \frac{3}{2}$; if $b>3$, then $a=\frac{b-\sqrt{b^2-9}}{2}$.

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Alternative solution.

Let $f(x)=\frac{a{x}^2+b}{\sqrt{x^2+1}}$. It is easy to see that $a>0$. Since ${\min }_{x\in \mathbb{R}}\frac{a{x}^2+b}{\sqrt{x^2+1}}=3$, and $f(0)=b$, we see that $b\ge 3$. $$f^{\prime }(x)=\frac{ax\left(x^2-\frac{b-2a}{a}\right)}{{\left(x^2+1\right)}^{3/2}}.$$

a. If $b-2a\le 0$, let $f^{\prime }(x)=0$, we have the solution $x_0=0$, and if $x<0$, then $f^{\prime }(x)<0$; and if $x>0$, then $f^{\prime }(x)>0$. $f(0)=b$ is the minimal value. Hence $b=3$, and $a\ge \frac{b}{2}$.

b. If $b-2a>0$, let $f^{\prime }(x)=0$, we have the solutions $x_0=0$, $x_{1,2}=\pm \sqrt{\frac{b-2a}{a}}$. It is easy to see $f(0)=b$ is not the minimal value, which implies $b>3$; and $f(x_{1,2})$ is the minimal value $$\begin{array}{rl}f(x_{1,2})& =\frac{a\cdot \frac{b-2a}{a}+b}{\sqrt{\frac{b-2a}{a}+1}}=3\Rightarrow 2\sqrt{a(b-a)}=3\\ \Rightarrow a^2-ab+\frac{9}{4}& =0\Rightarrow a=\frac{b-\sqrt{b^2-9}}{2},\end{array}$$ that is, $b>3$ and $a=\frac{b-\sqrt{b^2-9}}{2}$.

Summing up, we get (1) the range of $b$ is $[3,+\infty )$. (2) If $b=3$, then $a\ge \frac{3}{2}$; if $b>3$, then $a=\frac{b-\sqrt{b^2-9}}{2}$.