66th Belarusian Mathematical Olympiad

a) Show, for example, that the polynomials $$P(x)=4x^4+4x^3+1\text{and}Q(x)=x^4+4x+4$$ satisfy the condition. By Cauchy's inequality, for any real $x$ the following inequalities hold: $$x^4+x^4+x^4+1\ge 4\sqrt[4]{x^4\cdot x^4\cdot x^4\cdot 1}=4|x{|}^3,(1)$$ $$x^4+1+1+1\ge 4\sqrt[4]{x^4\cdot 1\cdot 1\cdot 1}=4|x|.(2)$$ From (1) it follows that $P(x)=4x^4+4x^3+1\ge x^4+4(|x{|}^3+x^3)\ge 0$, the last inequality turns to the equality only for $x=0$, but $P(0)=1$, so the polynomial $P(x)$ has no real roots. Similarly, from (2) it follows that $Q(x)=x^4+4x+4\ge 4(|x|+x)+1>0$, whence the polynomial $Q(x)$ has no real roots. By definition, $$P_Q(x)=4x^4+4x+1\text{и}{Q}_P(x)=x^4+4x^3+4$$ It is easy to verify that $P_Q(-0.5)=-0.75<0$ and $P_Q(0)=1>0$. Also, $Q_P(-2)=-12<0$ and $Q_P(0)=4>0$. Therefore each of the polynomials $P_Q(x)$ and $Q_P(x)$ has at least one real root.

b) It is evident that the polynomials $P(x)$ and $Q(x)$ satisfying the condition of a) must have even degree (since any odd degree polynomial has at least one real root). Show that the degree of these polynomials is greater than 2. Suppose, contrary to our claim, that there are the trinomials $P(x)=a_1{x}^2+b_{1}x+c_1$ and $Q(x)=a_2{x}^2+b_{2}x+c_2$ having no real roots. Then at least one of the trinomials $P_Q(x)=a_1{x}^2+b_{1}x+c_1$ and $Q_P(x)=a_2{x}^2+b_{2}x+c_2$ has no real roots. Indeed, without loss of generality we can assume that $|b_2|\ge |b_1|$. Since the discriminant of the trinomial $Q(x)$ is negative and $|b_2|\ge |b_1|$, we see that $0>b_2^2-4a_2{c}_2\ge b_1^2-4a_2{c}_2$. However, the last expression is a discriminant of the trinomial $Q_P(x)$, so this trinomial has no real roots. The example of the polynomials from item a) shows the minimal degree of the polynomials is equal to 4.