Belarusian Mathematical Olympiad

Let $x_1$, $x_2$ be distinct roots of the equation $a{x}^2+bx+c=0$, i.e. be zeroes of the functions $f(x)=a{x}^2+bx+c$ and $g(x)=x^3+b{x}^2+ax+c$. By condition, $f(0)=g(0)=c\ne 0$. Let $F(x)=g(x)-f(x)$. Then $0$, $x_1$, $x_2$ are the distinct zeroes of the polynomial $F(x)$. So $F(x)=x(x-x_1)(x-x_2)$, hence $$\begin{gathered} x^3+(b-a)x^2+(a-b)x=x(x-x_1)(x-x_2) \\ (b-a)x+(a-b)=-(x_1+x_2)x+x_1{x}_2\forall x\in \mathbb{R}. \end{gathered}$$ Therefore, $x_1+x_2=a-b$ and $x_1{x}_2=a-b$. By Vieta's theorem, $x_1+x_2=-b/a$, $x_1{x}_2=c/a$, which gives $c=b$. Further, $a-b=-b/a$ or $a^2-ab=-b$, i.e. $a^2=b(a-1)$. Then $a\ne 1$, and from $b>0$ it follows that $a>1$.

a) Since $b=a^2/(a-1)$, we have $$c=b=\frac{a^2}{a-1}=\frac{(a-1+1{)}^2}{a-1}=\frac{(a-1{)}^2+2(a-1)+1}{a-1}=(a-1)+2+\frac{1}{a-1}\ge 4.$$ Therefore, $abc=a{b}^2>1\cdot 4^2=16$, as required.

b) First solution. Since $ab=a^2+b$, we see that $$abc=a{b}^2=(a^2+b)b=\left(\frac{a^{2}b}{2}+\frac{a^{2}b}{2}+\frac{b^2}{3}+\frac{b^2}{3}+\frac{b^2}{3}\right)\ge 5\sqrt[5]{\frac{a^{2}b}{2}\cdot \frac{a^{2}b}{2}\cdot \frac{b^2}{3}\cdot \frac{b^2}{3}\cdot \frac{b^2}{3}}=5\sqrt[5]{\frac{a^4{b}^8}{4\cdot 27}},$$ hence $$(a{b}^2{)}^5\ge \frac{5^5}{108}{a}^4{b}^8⟹abc=a{b}^2\ge \frac{3125}{108},$$ as required. Note that the equality is achieved only if $\frac{a^2}{2}=\frac{b}{3}$, then from $b=\frac{a^2}{a-1}$ it follows that $a-1=2/3$, i.e. $a=5/3$.

Second solution. We have $abc=a{b}^2=\frac{a^5}{(a-1{)}^2}$. We find the smallest value of $h(a)=\frac{a^5}{(a-1{)}^2}$ on $(1;+\infty )$. We see that $$h^{\prime }(a)=0⟺\frac{5a^4(a-1{)}^2-2(a-1)a^5}{(a-1{)}^4}=\frac{a^4(a-1)(3a-5)}{(a-1{)}^4}=0,$$ so $a=5/3$. Since $h^{\prime }(a)<0$ on $(1;5/3)$ and $h^{\prime }(a)>0$ on $(5/3;+\infty )$, we see that the smallest value of $h$ on $(1;+\infty )$ is equal to $h(5/3)=\frac{5^5}{108}$. Therefore, $abc\ge \frac{5^5}{108}$, as required.