Update I 'repaired' this method, but it changed a lot and I have some different questions, so I posted it separately here.

As training for the olympiad, I have to solve a lot of inequalities. Recently, I found a very general method to solve inequalities. I think it's too good to be true, but if it is true, I think an implementation of the algorithm in a computer program could be used to very quickly prove all kinds of inequalities.

First, I'll explain the method. Then I'll ask my questions. At the very bottom, I proved a simple inequality using my method as sort of a demonstration.

The method / algorithm

Say we have to prove a certain inequality that looks like: $$f(a,b,c\ldots)\ge g(a,b,c\ldots)$$ for all $a,b,c\ldots\in \Bbb{R}$. Here $a,b,c\ldots$ are simply de independent variables. There might be just one, there might be $21$. It doesn't matter.

First, we rearrange: $$f(a,b,c\ldots)-g(a,b,c\ldots)\ge 0$$ Now, in order to prove our inequality, we have to prove that the LHS has a minimum value and that that minimum value is greater than or equal to $0$. Say that the inequality is true. Let $a_0,b_0,c_0\ldots$ be the values of $a,b,c\ldots$ for which the LHS is minimal. Now, we must have: $$\frac{d\,(f-g)}{d\, a}(a_0,b_0,c_0\ldots)=0$$ because otherwise, we could increase or decrease $a$ and get a lower value for the LHS, contradicting our definition of $a_0$. In fact, we have: \begin{align*} \frac{d\,(f-g)}{d\,a}(a_0,b_0,c_0\ldots) &= 0\\ \frac{d\,(f-g)}{d\,b}(a_0,b_0,c_0\ldots) &= 0\\ \frac{d\,(f-g)}{d\,c}(a_0,b_0,c_0\ldots) &= 0\\ &\vdots \end{align*} So we get as many equations as variables. Now we have a nice system of equations to work with. There are three options.

  1. The system has no solutions. This means a minimum does not exist and the inequality doesn't hold for all values of $a,b,c\ldots$.
  2. The system has exactly one solution. If we plug this in and also some other random values for $a,b,v\ldots$, we can see whether it's a maximum or a minimum. If it's a maximum, there are no minima, because the system only had one solution. If it's a minimum, simply plug in the values to check whether the inequality holds. If it does, it holds for all $a,b,c\ldots\in\Bbb{R}$.
  3. The system has infinitely many solutions. In this case, we can express these solutions using less variables than we had in the original inequality. Plugging in those solutions gives a new inequality with less variables, so if we repeat the proces, we eventually get a system of the first or second type or an inequality with just one variable and those are generally pretty easy to solve.


I have two questions.

  1. Does the method theoretically work? I think so, but there could be something I missed. As I said before, it looks too good to be true.
  2. If it does work, why don't people use it that often? I understand that proofs using things like AM-GM-HM are shorter, but this method is guaranteed to work. It either gives a full proof or a concrete counterexample.


Prove that for all $a,b,c\in\mathbb{R}$, we have: $$ab+bc+ca+a-c\le 1+\frac13(a+b+c)^2$$ Proof: First, rearrange: $$1+\frac13(a+b+c)^2-ab-bc-ca-a+c\ge 0$$ Now, take the derivatives with respect to $a$, $b$ and $c$ and set them equal to $0$: \begin{align*} \frac23(a+b+c)-b-c-1&=0\\ \frac23(a+b+c) -a-c&= 0\\ \frac23(a+b+c) - b - c + 1 &= 0 \end{align*} Some algebra to give us a nicer system: \begin{align*} b+c+3 &= 2a\\ a+c &= 2b\\ b+c-3 &= 2c \end{align*} So $c = 2b-a$. Plugging it into the first equation gives: \begin{align*} b + 2b - a + 3&= 2a\\ 3b + 3 &= 3a\\ b &= a-1 \end{align*} plugging that into the third equation gives: \begin{align*} a-1+c-3&=2c\\ a-2 &= c \end{align*} and that's the solution to our system; $b=a-1$, $c=a-2$. Plugging this into the original inequality gives: \begin{align*} ab+bc+ca+a-c &\le 1+\frac13(a+b+c)^2\\ a(a-1) + (a-1)(a-2) + (a-2)a + a - (a-2) &\le 1+ \frac13(3a-3)^2\\ a^2 - a + a^2 - 3a + 2 + a^2-2a + 2 &\le 1+\frac13(9a^2-18a+9)\\ 3a^2 - 6a + 4 &\le 1 + 3a^2 - 6a + 3\\ 3a^2 -6a + 4 &\le 3a^2 -6a + 4 \end{align*} Which always holds and therefore the original inequality always holds. Q.E.D.

  • $\begingroup$ Apart from the other comments, why should a system of equations be nice? There are lots of system of equations that we don't know how to even approximate in practice. $\endgroup$
    – mlainz
    Commented Apr 30, 2017 at 15:15
  • $\begingroup$ Another point that I haven't seen mentioned yet. Your main system of equations (partial derivatives set to zero) can have any non-negative integer number of solutions, 0, 1, 2, 3, 10, 162 etc.The system will only have 0, 1 or infinitely many real solutions if these derivatives are linear. The solution to a system on of nonlinear equations can yield all sorts of solutions. $\endgroup$ Commented May 1, 2017 at 12:30
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    $\begingroup$ Suggetion: Read about Lagrange multipliers and Lagrange methods for finding extrema, which you are in the proces of re-discovering. $\endgroup$ Commented May 2, 2017 at 0:08

3 Answers 3


Actually, this is one approach to doing inequalities, and sometimes it is used. But there are some caveats that need to be paid attention to:

  1. It is not always the case that those multi-variable functions are differentiable at all points that you are interested in. (And in some cases, the function is differentiable a.e., but the function is so bizarre that its derivative does not contain too much information about the function's variation - this is rare though, usually rises theoretically but not in practical use cases.)

  2. Even if 1 is not an issue, sometimes it is not easy to get the differentiation of the function - it could be tricky to calculate differential values for certain points, or the derivative is very involving for the function expressed in nested combinations of elementary operations/functions, or the derivative may not be in closed-form if the function is not elementary, etc. (Plus if the differentiation is in a complicated form, it is also difficult to solve the equations of which derivatives equal to zero.)

  3. Even if 2 is not an issue, it is not guaranteed that the point you get by having first-order partial derivatives equaling to zero is local minimum/maximum, you might just get a saddle point - thus second-order derivatives need to be checked as well (notice also there is no guarantee that second-order derivatives exist)

  4. Even if 3 is not an issue, you still cannot be sure that the local maximum/minimum is global maximum/minimum - e.g. it could be possible that points at the boundary are larger than local maximum point you get. So check more before you make the final conclusion - For example, when finding the global maximum:

    • Compare all local maximum points (if there is any, and make sure they are local max);
    • Compare all points at the boundary (if there is any);
    • Check function's behavior as it approaches infinity (if applicable);
    • Check where first-order derivatives do not exist (if any) - they could be maximum points;
    • and so on..
  • 6
    $\begingroup$ Thank you for your clear answer. I really should've been more careful when I 'created' this method. $\endgroup$
    – Mastrem
    Commented Apr 30, 2017 at 16:04
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    $\begingroup$ @Mastrem Thanks, it's a very good thought process by creating this question/method, so no worries! $\endgroup$
    – Jay Zha
    Commented Apr 30, 2017 at 16:14
  • $\begingroup$ But this method would surely work for all sorts of algebraic combinations of $a,b,c...$ right? (By algebraic combinations I mean to say all inequalities involving only integral powers of the variables in the terms). $\endgroup$ Commented May 1, 2017 at 14:42
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    $\begingroup$ @FreezingFire Yep, I would think most of the simple functions should work. But still pay attention, some of it looks simple, but might be tricky. For example $\frac{ab}{a^2+b^2}$ is not differentiable at $(0,0)$; or simple algebra (like $a^2-b^2$) could also result in saddle point, etc - but I general I would say it's not a big issue, as long as you pay attention to the points I listed above. $\endgroup$
    – Jay Zha
    Commented May 1, 2017 at 14:48

The answer by @yujie-zha misses a few important points, so let me elaborate on this.

Congratulations, you rediscovered a well-known theorem which is sometimes called in France “principe de Fermat,” sharing the same name than a reflection law in optics and the remark according to which there is no strictly decreasing sequence in positive numbers. Pierre de Fermat was a French mathematician living in the 17th century.

Your discussion covers a lot of useful cases but it has two major flaws:

  1. A point where the quantity $F = f - g$ reaches its minimum does not need to exist. A simple example is provided by $1/x$ for $x\gt0$.

  2. If such a point exists, the geometry of the set where the minimum of $F$ is realised can be arbitrarily complicated, so even if it has infinitely many points, in does not need to be so nice that you actually can remove a variable. Functions $F$ which allow to remove variables are called submersions. Furthermore, finding that point can be impracticable.

Regarding 1, a common hypothesis guaranteeing that the minimum is actually realised is when the function $F$ we study is continuous and defined over a compact domain. This might not tell you much for now, but you can realise what happen if you try to build a function always positive on the real line but whose minimum “flees to the infinity” – a colloquial way to say that it's asymptotically zero, if you are already familiar with these terms.

Regarding 2, a theorem by Borel states that any closed subset of an euclidean space is the zero locus of a smooth function. This will also probably not tell you much, but you can try to find functions $F$ that are positive, everywhere differentiable over the real line and whose zero locus is:

  • The set of integers.
  • The set of inverse of integers, to which we add the point $0$.
  • The segment $[-1,1]$.
  • The Cantor set (if you know this one).

You can also look for similar examples in higher dimensions.

This aside, the method you suggest actually works, but happens to be not as easy to use as you seem to think. Other common sources for inequality are convex functions, esp. Jensen's inequality, the Cauchy-Schwarz or Cauchy-Bouniakovski inequality, and the modulus of a holomorphic function.

  • 2
    $\begingroup$ The whole field of optimization and operations research would be tremendously simplified if you could just look at the zeroes of the derivative only, not also at all the places where the derivative doesn't exist, to find the minimum. $\endgroup$
    – Joker_vD
    Commented May 2, 2017 at 8:48

There are a few issues with this method.

  1. First, it's unclear whether you are trying to solve an inequality or prove an inequality. They're not the same thing: Proving an inequality means showing that it's true for all values of the variables (as for example in the inequality $(x+y)^2 \ge 4xy$), whereas solving an inequality means finding the values of the variables for which the inequality is true (as for example in the inequality $x^2 - 1 < 2x$, which is true for $x \in (1-\sqrt{2},1+\sqrt{2})$.

Of course it could be argued that part of solving an inequality always involves a proof, but still it is important to know whether you are trying to show that a given inequality holds for all values of the variable(s), or whether it holds for some values of the variable(s).

  1. You write:

Now, in order to prove our inequality, we have to prove that the LHS has a minimum value and that that minimum value is greater than or equal to 0

This is not true. For a simple counterexample, consider the inequality $e^{-x}>-2$, which is definitely true for all (real) $x$. But the function $f(x)=e^{-x}+2$ has no minimum value. It has a lower bound, even a greatest lower bound (namely $f(x) \ge 2$) but it never actually attains that greatest lower bound.

  1. Even if a function does attain a minimum value, there is no guarantee that the derivative there is $0$. For example, the function $f(x)=x^{2/3}$ is defined on all of $\mathbb{R}$ and attains its minimum value at $x=0$, but the derivative there is not $0$ because the function isn't differentiable at $x=0$. In general, remember that critical values of a function occur where the derivative is zero or undefined.

  2. If you are dealing with a function that is defined on some restricted domain (rather than on the whole real line), a minimum value might be attained at an endpoint, rather than at a critical value. For example if we consider the function $f(x)=\tan(x)$ defined on $[0, \pi/2)$ (which would be the reasonable domain if $x$ represents an angle in a right triangle) then the minimum occurs at $x=0$.

Having said all of this, it is true that if you can find the global minimum of a function, and show that minimum value is positive, that would establish the inequality you are trying to prove. But that's a pretty big "if".

  • $\begingroup$ Why pick $e^{-x}+2$ instead of $e^{-x}$? BTW Functions like $1/x$ for $x>0$ or $\frac{1}{1+x^2}$ for real $x$ are arguably easier to work with. $\endgroup$ Commented May 2, 2017 at 9:10
  • $\begingroup$ @MichaelGrünewald The OP turned $f>g$ into $f-g>0$, and thought that $f-g$ had to have a positive global minimum. To prevent confusion, I wanted an example with a nonzero $g$ (so that it would still look like "something minus something") and with a positive infimum. $\endgroup$
    – mweiss
    Commented May 2, 2017 at 13:05
  • $\begingroup$ @MichaelGrünewald In fact I initially planned to use $1/x$, but rejected it on the grounds that it's not defined on all of $\mathbb{R}$, which introduces an additional fact that's not really relevant for the question but could be distracting. $\frac{1}{1+x^2}$ would have been fine, of course, but I'm not sure why you think it's "easier" to work with. $\endgroup$
    – mweiss
    Commented May 2, 2017 at 13:07
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    $\begingroup$ $\frac{1}{1+x^2}$ is easy to define, the exponential function is harder. Since the OP seems to be discovering calculus, they may be familiar or not with the exponential function and rational functions might be a safer pedagogical tool. $\endgroup$ Commented May 3, 2017 at 18:56
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    $\begingroup$ @MichaelGrünewald Not that it matters, but a few observations: (1) "easy to define" and "easy to understand" are two different things. (2) In the United States, exponential functions are encountered well before Calculus -- they are part of the Algebra 2 curriculum and are a major theme of Precalculus. (3) The OP says that he/she is training for the Olympiad; it seems unlikely to me that such a student would not be familiar with exponential growth and decay. $\endgroup$
    – mweiss
    Commented May 3, 2017 at 19:00

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