Prove that there exists a polynomial $P(x)$ with real coefficients and degree greater than 1 such that both of the following conditions are true $\cdot$ $P(a)+P(b)+P(c)\ge 2021$ holds for all nonnegative real numbers $a,b,c$ such that $a+b+c=3$ $\cdot$ There are infinitely many triples $(a,b,c)$ of nonnegative real numbers such that $a+b+c=3$ and $P(a)+P(b)+P(c)= 2021$
Problem
Source: Thailand Online MO 2021 P5
Tags: algebra, polynomial, Thailand online MO
Mahdi.sh
05.04.2021 16:18
Prove that there exists a polynomial $P(x)$ with real coefficients and degree greater than 1 such that both of the following conditions are true $\cdot$ $P(a)+P(b)+P(c)\ge 2021$ holds for all nonnegative real numbers $a,b,c$ such that $a+b+c=3$ $\cdot$ There are infinitely many triples $(a,b,c)$ of nonnegative real numbers such that $a+b+c=3$ and $ P(a)+P(b)+P(c)= 2021 $
hsgsstudent
16.09.2021 07:23
we prove $P(x)=x^{3}-9x^{2}+18x+\dfrac{2021}{3}$ satisfies problem. we have
$$\begin{aligned}
&P(a)+P(b)+P(c)=P(a)+P(b)+P(3-b-a)\\
&=a^{3}-9a^{2}+b^{3}-9b^{2}+(3-b-a)^{3}-9(3-b-a)^{2}+2021+18(a+b+c)\\
&=\left[a^{3}+b^{3}+(3-b-a)^{3}\right]-9\left[a^{2}+b^{2}+(3-b-a)^{2}\right]+2021+54\\
&=\left[27-27(a+b)+9(a+b)^{2}-3ab(a+b)\right]-9\left[9-6(b+a)+2(b+a)^{2}-ab\right]+2021+54\\
&=27(a+b)-9(a+b)^{2}+3ab\left[3-(a+b)\right]+2021\\
&=\left[9(a+b)+3ab\right]\left[3-(a+b)\right]+2021
\end{aligned}$$Since $a,b,c \ge 0$ and $a+b+c=3$ , we have $9(a+b)+3ab \ge 0$ and $a+b \le 3$
So $\left[9(a+b)+3ab\right]\left[3-(a+b)\right] \ge 0 \Rightarrow P(a)+P(b)+P(c) \ge 2021$
Equality hold if $a+b=3$ or $(a,b,c)=(x,3-x,0)$ with $x \in \left[0,3\right]$