socrates wrote: Find all positive integers $n$ for which $$3x^n + n(x + 2) - 3 \geq nx^2$$holds for all real numbers $x.$ It's wrong! Try $n=3$ and $x=-100991$. It was my delirium, of course.

arqady wrote: socrates wrote: Find all positive integers $n$ for which $$3x^n + n(x + 2) - 3 \geq nx^2$$holds for all real numbers $x.$ It's wrong! Try $n=3$ and $x=-100991$. Is it a problem of finding a $n$?

It's equivalent to $3(x^n-1)\ge n(x^2-x-2)$. If $n$ is odd then take $x=-1$. Then $3(x^n-1)=-6$ but $n(x^2-x-2)=0$ so it doesn't hold.

The inequality holds for every even positive integer $n$.Plugging in $x=-1$ we get $3(-1)^n -3 \geq 0$,which implies that $n$ is even.Suppose $ x \geq 0 $.$x=0,1$ clearly satisfy the inequality so we may assume $x \geq 2$.For $n=2$ the inequality reduces to $(x+1)^2 \geq 0$,which is obviously true.So,suppose $n \geq 4$. We have: $3x^n -nx^2+nx+2n-3 \geq 0$.So,it would suffice to show that :$3x^n \geq nx^2 $.This can be proven easily with induction using the fact that $x \geq 2$ and $n \geq 4$.Suppose $x \leq0$.Again $x=-1$ satisfies the inequality so we may assume $x \leq -2$.It would suffice to prove that $3x^n-nx^2+nx \geq 0$.Which implies that :$3x^{n-1} -nx +n \leq 0$.Substitute $y=-x$.Using the fact that $n$ is even,we get to prove that :$3y^{n-1} \geq n(y+1)$,which can again be proven by induction and using the facts $y \geq 2$,$n \geq 4$.

Our solution: All even $n$. It is easy to see that the inequality does not hold for odd $n$. For even $n$ use AM-GM to get \[ x^{n}+\underbrace{1+1+\dots+1}_{(n-1)- \text{times}} \geq n \sqrt[n]{x^{n}} = n|x| \geq -nx \]and \[ x^{n}+x^{n}+\underbrace{1+1+\dots+1}_{(n-2)-\text{times}} \geq n \sqrt[n]{x^{n} \cdot x^{n}} = n x^2. \]Note that it's valid to use AM-GM. Adding yields the conclusion.

wonderful

Plugging in $x=-1$ gives $n$ even. For $n=2$ it is $x^2+2x+1 \geq 0$ which is obvious. If $n \geq 4$. Let $n=2k$. The inequality is equivalent to $$ (x+1)^2 \left( 3(x-1)^2 \left( \sum_{i=1}^{k-1} ix^{2k-2i-2} \right) + k \right) \geq 0$$which is also obvious.