9
$\begingroup$

Let $f(x)=x^4+ax^3+3x^2+bx+1$, where $a,b \in \mathbb{R}$. If $f(x) \ge 0$ for all $x \in \mathbb{R}$, what is the maximum possible value of $a^2+b^2$?

I don't know how to proceed. Hints or help will be appreciated.

  • 0
    For negative values in $R$ , each of these negative values $ax^{3}$,$bx $add upto a non-negative quantity $a$ , $b$ can also be negative too> , can we say something from that? – 2017-02-11
  • 0
    $$\forall x:f(x)\geq0 $$so I think $f(x)$ must be in form of $(x-k_1)^4$ or $(x-k_1)^2(x-k_2)^2$ – 2017-02-11
  • 0
    @Khosrotash Why? $f(x)=x^4+3x^2+1 \gt 0$ and it's in neither of those forms. – 2017-02-11
  • 0
    @Shobhit I guess you mean something easier than actually calculating the discriminant of the quartic and checking [all subcases](https://en.wikipedia.org/wiki/Quartic_function#Nature_of_the_roots). – 2017-02-11
  • 0
    @dxiv its a contest question, there must be a catch. – 2017-02-11
  • 0
    @Shobhit Which is precisely why it's advised to provide more context to the question. – 2017-02-11
  • 0
    @Shobhit: The problem is from what contest? – 2017-02-11
  • 0
    This seems to be basically the same question as [Polynomial maximization: If $x^4+ax^3+3x^2+bx+1 \ge 0$, find the maximum value of $a^2+b^2$](http://math.stackexchange.com/q/1169302). I found it [using Approach0](https://approach0.xyz/search/?q=%24x%5E4%2Bax%5E3%2B3x%5E2%2Bbx%2B1%5Cge0%24&p=1). (We will see whether it will also appear among related questions, now that title has been changed to be more descriptive.) – 2017-02-11

2 Answers 2

3

$$f(x) = \frac{1}{4}\bigg[4x^4+4ax^3+12x^2+4bx+4\bigg]\geq 0$$

$$f(x)= \frac{1}{4}\bigg[(2x^2+ax)^2+(bx+2)^2+(12-a^2-b^2)x^2\bigg]\geq 0$$

So $$12-(a^2+b^2)\geq 0.$$

So $$a^2+b^2\leq 12$$

  • 7
    @Shobhit : This is not correct. $a^2+b^2=13$ is possible since $x^4+0\cdot x^3+3x^2+\sqrt{13}\ x+1$ is [non-negative](http://www.wolframalpha.com/input/?i=x%5E4%2B3x%5E2%2B(sqrt(13))*x%2B1). – 2017-02-11
  • 0
    A fault lies in going from $c^2+d^2+ke^2 \geq 0$ to conlude that $k \geq 0$. This conclusion could be drawn if we had $-c^2-d^2+ke^2 \geq 0$ instead, but this is not the case. – 2017-02-11
  • 0
    @mathlove i see your point. then if this is not correct, how to solve it then? Still, i would like to point out, that the given answer is 12 (textbook) – 2017-02-11
  • 0
    @DURGESH TIWARI You proof is not full. Why does the equality occur? – 2017-02-11
  • 0
    @DurgeshTiwari The answer is wrong, as hinted in a previous comment and proved at [the duplicate question](http://math.stackexchange.com/questions/1169302/polynomial-maximization-if-x4ax33x2bx1-ge-0-find-the-maximum-value-o), so I suggest you retract it, lest future readers might get confused by the upvotes. – 2017-02-15
  • 0
    yes dxive you are right. – 2017-02-15
2

Following mathlove's strategy, we can do a little better . . .

If $f(x) = x^4 + ax^3 + 3x^2 + bx + 1$, where

\begin{align*} a&=0\\[8pt] b&=\sqrt{6+{\small{\frac{14}{9}}}\sqrt{21}} \end{align*}

then $f(x) \ge 0$ for all $x \in \mathbb{R}$, and

$$a^2 + b^2= 6+{\small{\frac{14}{9}}}\sqrt{21} \approx 13.12845108$$

I don't know if this is best possible, however it is best possible for the case $a=0$.