In this notebook there are presented examples of usage of shiroin, a python library for proving inequalities of multivariate polynomials.

At the beginning we need to load the packages.

shiro.seed=1 sets a seed for proving functions. If you don't write it, you can get a slightly different proof each time you run a function. This line is here only for the sake of reproducibility.

The next line provides a nicer display of proofs, i.e. formulas will be shown instead of LaTeX code of these formulas. Note that this works on Jupyter, but not on the git page.

Now let's make some proofs. We will use problems from https://www.imomath.com/index.php?options=593&lmm=0.

Problem 1¶

Prove the inequality $a^2+b^2+c^2\ge ab+bc+ca$, if $a,b,c$ are real numbers.

Function prove tries to prove that given formula is nonnegative, assuming all variables are nonnegative. In this case the nonnegativity assumption is not a problem, since all powers on the left side are even, so if $|a|^2+|b|^2+|c|^2 \ge |ab|+|ac|+|bc|,$ then $a^2+b^2+c^2= |a|^2+|b|^2+|c|^2 \ge |ab|+|ac|+|bc| \ge ab+ac+bc$.

Function prove prints several things. The first two gives us a formula after expanding it. The next one is status, which is the return status of the first use of scipy.optimize.linprog. Possible outputs and explanations are

• 0 - found a proof with real coefficients,
• 1 - need more time,
• 2 - function didn't find a proof,
• 3,4 - loss of precision (which may happen if it has to work with big numbers).

Then we've got a hint. So let's use it!

Problem 2¶

Find all real numbers such that $a^2+b^2+c^2+d^2=a(b+c+d)$.

At first glance it doesn't look like an inequality problem, but actually it is one. If you try to calculate both sides for different values, you can see that the left side of the equation is never less than the right one. So let's try

This time prove didn't found the proof. But it doesn't mean that the inequality is not true! prove uses a list of $n$ positive values, where $n$ is a number of variables in the formula. List of values should correspond to the list of variables in alphabetical order. Here are a few tips how to choose the right values.

1. Consider a function $pos(values)$ which is the sum of the positive addends in the formula after substitution of values to variables. Analogically, let $neg(values)$ be the sum of the negative addends. We should choose such values for which $quotient=pos(values)/|neg(values)|$ is small.
2. The symmetry group of the formula splits set of variables into orbits. Using the same values for variables in one orbit is recommended. In particular, if the symmetry group of the formula is transitive (for example, when the formula is cyclic), then all values (probably) should be the same.
3. If the formula is homogeneous, then $values=(a_1,a_2,...,a_n)$ provide a proof iff $values=(ka_1,ka_2,...,ka_n)$ provides a proof for any $k\in Q_+$ (as long as you don't run into overflow error).

In the formula above $b,c,d$ are in one orbit and the formula is homogenous, so let's try $a=2$ and $b=c=d=1$.

Function makes a substitution $a\to 2e$ and try to prove new inequality. This time it succeeded. Moreover, if starting formula is equal to 0, then all these inequalities have to be equalities, so $e^2=0$ and eventually $a=0$. We can also try a little bit lower value for $a$.

Now we can see that if $a^2+b^2+c^2+d^2-a(b+c+d)=0$, then $7f^2+2b^2+2c^2+2d^2=0$ and eventually $a=b=c=d=0$. Note that inequality is proved only for positive numbers (which, by continuity, can be expanded to nonnegative numbers). But using similar argumentation to the one in previous problem, if $(a,b,c,d)=(x,y,z,t)$ is the solution of $a^2+b^2+c^2+d^2-a(b+c+d)=0$, then $(a,b,c,d)=(|x|,|y|,|z|,|t|)$ is a solution, too. Since the only nonnegative solution is $(0,0,0,0)$, it means that it is the only solution.

It is worth noting that this time function prove used $f$ as a new variable instead of $e$. If you want to start a new proof and you don't care about the collision of variables from previous proofs, you can use newproof function, which clears the set of used variables.

We can also use the findvalues function to find values for the proof more automatically. It looks for (local) minimum of the $quotient$ value defined above.

If the $quotient$ value were less than 1, that would mean that the formula is negative for given values. If $quotient$ were equal to 1, then we have to choose exactly these values (or other values for which the $quotient$ is equal to 1. But here $quotient$ is greater than 1, so we can take a point near it and (probably) still have a proof. The values given to the prove function must not be floating point numbers, so we can rationalize them.

If you set the tolerance bigger, then the values will have smaller numerators and denominators, so coefficients in the proof will be smaller, too. But if the tolerance is too big, then proof will not be found.

Let's skip the problem 3 and look solve the problem 4 instead.

Problem 4¶

If $x$ and $y$ are two positive numbers less than 1, prove that

prove assumes that formula is well-defined if all variables are positive, so it doesn't have to analyze the denominator (except of choosing the right sign). In this case it is not true, since if $x=1$, then $1-x^2=0$. Also denominator is equal to $(x^2-1)(y^2-1)(xy-1)$ which is negative for $x,y\in (0,1)$. So we need to make some substitution after which new variables can have all positive values, not just these inside (0,1) interval.

We will use a function makesubs to generate these substitutions. It has three basic parameters: formula, intervals and values. intervals are current limitations of variables, values are values of variables for which $quotient$ of formula is small. values should be inside corresponding intervals. This argument is optional but it's better to use it. Let's go back to our problem. If $x=y$, then $\frac{1}{1-x^2}+\frac{1}{1-y^2}\ge \frac{2}{1-xy}$, so it's the minimum value of the formula. So let values=(1/2,1/2) (warning: do not use decimal point, for example '0.5,0.5').

Now let's get back to problem 3.

Problem 3¶

If $a,b,c$ are positive real numbers that satisfy $a^2+b^2+c^2=1$, find the minimal value of

The problem is equivalent to finding minimum of $xy/z+yz/x+zx/y$ assuming $x+y+z=1$ and $x,y,z>0$. The first idea is to suppose that the minimum is reached when $x=y=z$. In that case, $x=y=z=1/3$ and formula is equal to 1. Now we can substitute $z\to 1-x-y$. Constraints for variables are $x>0$, $y>0$, $x+y<1$. We can rewrite it as $x \in (0,1-y)$, $y \in (0,1)$. These two conditions have two important properties:

• constraints for variables are written as intervals,
• there are no "backwards dependencies", i.e. there is no $x$ in the interval of $y$.

If these two conditions hold, then you can use makesubs function.

The proof is found, so the assumption that 1 is the minimum of xy/z+yz/x+zx/y was good.

Functions S and Sm creates a SymPy object from a string. The only difference is that Sm assumes that there are no multi-letter variables and adds a multiplication sign between every two terms which has no operator sign, so object Sm(xy/z+yz/x+zx/y) has 3 variables x,y,z and S('xy/z+yz/x+zx/y') has 6 variables x,y,z,xy,yz,zx.

As you may have noticed, formulas are often cyclic or symmetric. Therefore you can use cyclize or symmetrize function to reduce the length of the written formula. Here are a few commands which will do the same as each other.

Now look at formula $(x-1)^4$. It's quite obvious that it's nonnegative, but prove fails to show this!

But there is a relatively simple method to generate a proof using this library. We will make to proofs: one for $x\in (1,\infty)$ and the second one for $(-\infty,1)$.

Now let's go to the problem 10

Problem 10¶

If $a,b,c,d>0$, prove that

Let's try a simple approach.

This problem, like the previous one, can be solved by splitting the domain of variables to several subdomains. But we can also use the symmetry of this inequality. For example, without loss of generality we can assume that $a\ge c$ and $b\ge d$, so $a\in [c,\infty)$, $b\in [d,\infty)$.

It's a good idea to use intervals that are unbounded from one side (i.e. those which contain $\pm\infty$). In this problem we could assume that $a\in (0,c]$, $b\in (0,d]$ as well. But as you can see, in this case the proof is several times longer.

Function powerprove is a shortcut for splitting domain $R_+^n$ on several subdomains and proving the inequality for each of them. This function uses $2^n$ of $n$-dimensional intervals with a common point (by default it's $(1,1,...,1)$), where $n$ is a number of variables. Here there are two examples of using it. As you can see, proofs are very long.

Now let's take a look at slightly another kind of the problem.

Problem¶

Let $f:R^3\to R$ be a convex function. Prove that

To create a proof, we will use provef function. It assumes that $f$ is convex and nonnegative, then it tries to find a proof. However, if the last inequality is $0\le 0$, then the proof works for any convex function.

Let's try to solve problem 6 from the finals of LXIII Polish Mathematical Olympiad. It was one of the hardest inequality in the history of this contest, solved only by 3 finalists.

Problem¶

Prove the inequality

The first observation is that the formula is cyclic, so without loss of generality we may assume that $a\ge b,c$. We can go a step further and divide it into two cases: $a\ge b\ge c$ and $a\ge c\ge b$.

So the case $a\ge c\ge b$ is done, but $a\ge b\ge c$ is not. But maybe we can adjust values.

First and second value is approximately equal to 0, so we can try to replace 0 with 1.

The key observation is that the formula1 is homogenous, so we can scale values.

Now the third value is very big. Technically we could use it, but it would run into overflow error, so we will use 1 instead of it. Second value is very close to $\sqrt{2}$, so this value will be our next try.

So after getting the code all together we have got the following proof.