This question is related to this one. Given two functions $f,g\colon \mathbb{R}^n \to \mathbb{R}$, and given that at a certain point $x_0 \in \mathbb{R}^n$ we have $f(x_0) \geq g(x_0)$, and given an open subset $A \subseteq \mathbb{R}^n$ containing $x_0$, on what conditions on the partial derivatives of $f$ and $g$ on $A$ can I assure that $f(x) \geq g(x)$ for every $x \in A$? When is this inequality strict?

This question came when I was trying to prove that for all $(x,y) \in \mathbb{R}^2$ and $(x,y) \neq (0,0)$, it holds that $x^2 + y^2 > xy$. Somehow, it seemed visually obvious that the function $x^2 + y^2$ grew faster than $xy$, and so they could never catch up again after the origin, but I found difficulties when trying to formalize this thought.

Also, is there a name for this result/test/condition, even if only on the one variable case?

  • $\begingroup$ My roommate just pointed out: the case where $xy < 0$ is trivial. For the other one, wlog, suppose $|x| \geq |y|$. Then, $x^2 \geq xy$, from which the inequality follows. I'm still curious about the derivative test, though. $\endgroup$ Commented Feb 9, 2017 at 19:18
  • $\begingroup$ Cant you simply use AM GM Inequality? Or more simply $ (x-y)^2$ is non negative $\endgroup$ Commented Jun 14, 2017 at 16:32
  • $\begingroup$ @GautamShenoy That would work for part 2, but doesn't answer part 1 :( $\endgroup$ Commented Jun 14, 2017 at 16:33
  • 1
    $\begingroup$ For part 2, note that $$x^2+y^2=\frac{x^2+y^2}2+\frac{(x-y)^2}2+xy\ge xy$$ $\endgroup$
    – san
    Commented Jun 16, 2017 at 15:39

1 Answer 1


While dodging your actual question, I will provide you this explanation to prove that $x^2+y^2\ge xy$.

We know that the root-mean square is bigger than or equal to geometric mean. $$\sqrt{\frac{x^2+y^2}{2}} \ge \sqrt{xy}$$ Clearly, $$x^2+y^2\ge 2xy > xy$$ assuming $x$ and $y$ are positive.

To give a small idea for your question, I can say if $f$ is bigger than $g$ at a point and then if the derivative of $f$ is bigger than derivative of $g$ everywhere (actually, after that point is enough), then you can conclude $f$ is bigger than $g$ everywhere after that point.

  • $\begingroup$ Are you sure about your last statement ? Let $f(x)=x,g(x)=x/2$, then $f'>g'$ and $f(1)>g(1)$ but $f(x<0)<g(x<0)$ $\endgroup$ Commented Jun 14, 2017 at 22:16
  • $\begingroup$ Because you started comparing them at $1$ and ensured that $f$ is bigger but at the end compared them for numbers smaller than $1$. Notice the last words: everywhere AFTER that point. I meant right side by "after". Sorry for lack of clarification $\endgroup$
    – user404177
    Commented Jun 14, 2017 at 22:38

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