25 KiB
from shiroindev import *
sSeed=1/home/grzegorz/shiroindev/shiroindev.py:40: DeprecationWarning: invalid escape sequence \^
formula=re.sub('\^{(.)}',r'^\1',latex(formula,fold_short_frac=True).replace(' ','').replace('\\\\left(','(').replace('\\\\right)',')'))
/home/grzegorz/shiroindev/shiroindev.py:41: DeprecationWarning: invalid escape sequence \\{
return re.sub('\\{(\(.+?\))\\}',r'\1',formula)
The first line obviously loads this package. The second one sets a seed for proving functions. If you don't write it, you can get slightly different proof each time you run a function.
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 $a^2+b^2+c^2= |a|^2+|b|^2+|c|^2 \ge |ab|+|ac|+|bc| \ge ab+ac+bc$.
prove('(a^2+b^2+c^2-a*b-a*c-b*c)')numerator: $$a^2-ab-ac+b^2-bc+c^2$$ denominator: $$1$$ status: 0 From weighted AM-GM inequality: Program couldn't find a solution with integer coefficients. Try to multiple the formula by some integer and run this function again. $$ ab+ac+bc \le a^2+b^2+c^2 $$
0
Function prove prints several things. The first two gives us a formula after expanding it. To proceed, a numerator has to be a polynomial with integer coefficients. 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!
prove('(a^2+b^2+c^2-a*b-a*c-b*c)*2')numerator: $$2a^2-2ab-2ac+2b^2-2bc+2c^2$$ denominator: $$1$$ status: 0 From weighted AM-GM inequality: $$2ab \le a^2+b^2$$ $$2ac \le a^2+c^2$$ $$2bc \le b^2+c^2$$ $$ 0 \le 0 $$ The sum of all inequalities gives us a proof of the inequality.
0
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
prove('a^2+b^2+c^2+d^2-a*(b+c+d)')numerator: $$a^2-ab-ac-ad+b^2+c^2+d^2$$ denominator: $$1$$ status: 2 Program couldn't find any proof. $$ ab+ac+ad \le a^2+b^2+c^2+d^2 $$
2
This time prove didn't found the proof. But it doesn't mean that the inequality is not true! prove uses a list of values for which the formula should be small. There is no strict rule here, but the smaller that value is, the higher are chances to find a proof. List of values should correspond to the list of variables in alphabetical order. So let's try $a=2$ and $b=c=d=1$.
prove('a^2+b^2+c^2+d^2-a*(b+c+d)','2,1,1,1')Substitute $ a \to 2a $ numerator: $$4a^2-2ab-2ac-2ad+b^2+c^2+d^2$$ denominator: $$1$$ status: 0 From weighted AM-GM inequality: $$2ab \le a^2+b^2$$ $$2ac \le a^2+c^2$$ $$2ad \le a^2+d^2$$ $$ 0 \le a^2 $$ The sum of all inequalities gives us a proof of the inequality.
0
Function makes a substitution $a\to 2a$ (which should be understood as $a=2a'$) 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 $a'^2=0$ and eventually $a=0$. We can also try a little bit lower value for $a$.
prove('a^2+b^2+c^2+d^2-a*(b+c+d)','7/4,1,1,1')Substitute $ a \to 7a/4 $ numerator: $$49a^2-28ab-28ac-28ad+16b^2+16c^2+16d^2$$ denominator: $$16$$ status: 0 From weighted AM-GM inequality: $$28ab \le 14a^2+14b^2$$ $$28ac \le 14a^2+14c^2$$ $$28ad \le 14a^2+14d^2$$ $$ 0 \le 7a^2+2b^2+2c^2+2d^2 $$ The sum of all inequalities gives us a proof of the inequality.
0
Now we can see that if $a^2+b^2+c^2+d^2-a(b+c+d)=0$, then $7a'^2+2b^2+2c^2+2d^2=0$ 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.
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 $$\frac{1}{1-x^2}+\frac{1}{1-y^2}\ge \frac{2}{1-xy}.$$
prove('1/(1-x^2)+1/(1-y^2)-2/(1-x*y)')numerator: $$-x^3y+2x^2y^2-x^2-xy^3+2xy-y^2$$
denominator: $$x^3y^3-x^3y-x^2y^2+x^2-xy^3+xy+y^2-1$$
status: 2
Program couldn't find any proof.
$$ x^3y+x^2+xy^3+y^2 \le
2x^2y^2+2xy $$
It looks like the formula is symmetric. You can assume without loss of generality that x >= y Try
prove(makesubs(S(" -x**3*y + 2*x**2*y**2 - x**2 - x*y**3 + 2*x*y - y**2 "), [('y', 'inf')] )
2
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 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').
newformula,newvalues=makesubs('1/(1-x^2)+1/(1-y^2)-2/(1-x*y)','[0,1],[0,1]','1/2,1/2')
prove(newformula*3,newvalues)Substitute $ x \to 1-1/(x+1) $ Substitute $ y \to 1-1/(y+1) $ numerator: $$6x^3y+3x^3-12x^2y^2-3x^2y+3x^2+6xy^3-3xy^2-6xy+3y^3+3y^2$$ denominator: $$4x^2y+2x^2+4xy^2+8xy+3x+2y^2+3y+1$$ status: 0 From weighted AM-GM inequality: $$12x^2y^2 \le 6x^3y+6xy^3$$ $$3x^2y \le 2x^3+y^3$$ $$3xy^2 \le x^3+2y^3$$ $$6xy \le 3x^2+3y^2$$ $$ 0 \le 0 $$ The sum of all inequalities gives us a proof of the inequality.
0
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 $$\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}$$
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.
Warning: at this moment makesubs doesn't warn you if your list of intervals doesn't follow these rules!
formula=Sm('xy/z+yz/x+zx/y-1').subs('z',S('1-x-y'))
newformula,values=makesubs(formula,'[0,1-y],[0,1]','1/3,1/3')
prove(newformula,values)Substitute $ x \to -y+1+(y-1)/(x+1) $ Substitute $ y \to 1-1/(y+1) $ Substitute $ y \to y/2 $ numerator: $$x^4y^2+x^3y^2-2x^3y-4x^2y+4x^2+xy^2-2xy+y^2$$ denominator: $$x^3y^2+2x^3y+2x^2y^2+4x^2y+xy^2+2xy$$ status: 0 From weighted AM-GM inequality: $$2x^3y \le x^4y^2+x^2$$ $$4x^2y \le x^3y^2+2x^2+xy^2$$ $$2xy \le x^2+y^2$$ $$ 0 \le 0 $$ The sum of all inequalities gives us a proof of the inequality.
0
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.
prove('(a^2+b^2+c^2-a*b-a*c-b*c)*2')
#prove(S('(a^2+b^2+c^2-a*b-a*c-b*c)*2'))
#prove(Sm('2(a^2+b^2+c^2-ab-ac-bc)'))
#prove(cyclize('2*a^2-2*a*b'))
#prove(symmetrize('a^2-a*b'))numerator: $$2a^2-2ab-2ac+2b^2-2bc+2c^2$$ denominator: $$1$$ status: 0 From weighted AM-GM inequality: $$2ab \le a^2+b^2$$ $$2ac \le a^2+c^2$$ $$2bc \le b^2+c^2$$ $$ 0 \le 0 $$ The sum of all inequalities gives us a proof of the inequality.
0
Now look at formula $(x-1)^4$. It's quite obvious that it's nonnegative, but prove fails to show this!
prove('(x-1)^4')numerator: $$x^4-4x^3+6x^2-4x+1$$ denominator: $$1$$ status: 2 Program couldn't find any proof. $$ 4x^3+4x \le x^4+6x^2+1 $$
2
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 $(-\infty,1)$.
prove(makesubs('(x-1)^4','(1,inf)'))Substitute $ x \to x+1 $ numerator: $$x^4$$ denominator: $$1$$ status: 0 $$ 0 \le x^4 $$ The sum of all inequalities gives us a proof of the inequality.
0
prove(makesubs('(x-1)^4','(-inf,1)'))Substitute $ x \to 1-x $ numerator: $$x^4$$ denominator: $$1$$ status: 0 $$ 0 \le x^4 $$ The sum of all inequalities gives us a proof of the inequality.
0
Now let's go to the problem 10
Problem 10
If $a,b,c,d>0$, prove that $$\frac a{b+c}+\frac b{c+d}+ \frac c{d+a}+ \frac d{a+b}\geq 2.$$
Let's try a simple approach.
formula=cyclize('a/(b+c)',variables='a,b,c,d')-2
formulaprove(formula)numerator: $$a^3c+a^3d+a^2b^2-a^2bd-2a^2c^2-a^2cd+a^2d^2+ab^3-ab^2c-ab^2d-abc^2+ac^3-acd^2+b^3d+b^2c^2-2b^2d^2+bc^3-bc^2d-bcd^2+bd^3+c^2d^2+cd^3$$ denominator: $$a^2bc+a^2bd+a^2c^2+a^2cd+ab^2c+ab^2d+abc^2+2abcd+abd^2+ac^2d+acd^2+b^2cd+b^2d^2+bc^2d+bcd^2$$ status: 2 Program couldn't find any proof. $$ a^2bd+2a^2c^2+a^2cd+ab^2c+ab^2d+abc^2+acd^2+2b^2d^2+bc^2d+bcd^2 \le a^3c+a^3d+a^2b^2+a^2d^2+ab^3+ac^3+b^3d+b^2c^2+bc^3+bd^3+c^2d^2+cd^3 $$
2
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 $a\in [c,\infty)$, $b\in [d,\infty)$.
prove(makesubs(formula,'[c,inf],[d,inf]'))Substitute $ a \to a+c $ Substitute $ b \to b+d $ numerator: $$a^3c+a^3d+a^2b^2+a^2bd+a^2c^2+2a^2cd+a^2d^2+ab^3+ab^2c+2ab^2d-abc^2+abd^2+b^3c+b^3d+b^2c^2+2b^2cd+b^2d^2$$ denominator: $$a^2bc+a^2bd+a^2c^2+2a^2cd+a^2d^2+ab^2c+ab^2d+3abc^2+6abcd+3abd^2+2ac^3+6ac^2d+6acd^2+2ad^3+b^2c^2+2b^2cd+b^2d^2+2bc^3+6bc^2d+6bcd^2+2bd^3+c^4+4c^3d+6c^2d^2+4cd^3+d^4$$ status: 0 From weighted AM-GM inequality: Program couldn't find a solution with integer coefficients. Try to multiple the formula by some integer and run this function again. $$ abc^2 \le a^3c+a^3d+a^2b^2+a^2bd+a^2c^2+2a^2cd+a^2d^2+ab^3+ab^2c+2ab^2d+abd^2+b^3c+b^3d+b^2c^2+2b^2cd+b^2d^2 $$
0
prove(makesubs(formula,'[c,inf],[d,inf]')*2)Substitute $ a \to a+c $ Substitute $ b \to b+d $ numerator: $$2a^3c+2a^3d+2a^2b^2+2a^2bd+2a^2c^2+4a^2cd+2a^2d^2+2ab^3+2ab^2c+4ab^2d-2abc^2+2abd^2+2b^3c+2b^3d+2b^2c^2+4b^2cd+2b^2d^2$$ denominator: $$a^2bc+a^2bd+a^2c^2+2a^2cd+a^2d^2+ab^2c+ab^2d+3abc^2+6abcd+3abd^2+2ac^3+6ac^2d+6acd^2+2ad^3+b^2c^2+2b^2cd+b^2d^2+2bc^3+6bc^2d+6bcd^2+2bd^3+c^4+4c^3d+6c^2d^2+4cd^3+d^4$$ status: 0 From weighted AM-GM inequality: $$2abc^2 \le a^2c^2+b^2c^2$$ $$ 0 \le 2a^3c+2a^3d+2a^2b^2+2a^2bd+a^2c^2+4a^2cd+2a^2d^2+2ab^3+2ab^2c+4ab^2d+2abd^2+2b^3c+2b^3d+b^2c^2+4b^2cd+2b^2d^2 $$ The sum of all inequalities gives us a proof of the inequality.
0
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.
prove(makesubs(formula,'[0,c],[0,d]')*2)Substitute $ a \to c-c/(a+1) $ Substitute $ b \to d-d/(b+1) $ numerator: $$2a^3bc^2d^2+4a^3bcd^3+2a^3bd^4+2a^3c^2d^2+2a^3cd^3-2a^2b^2c^3d+2a^2b^2cd^3-4a^2bc^3d+4a^2bc^2d^2+12a^2bcd^3+6a^2bd^4-2a^2c^3d+4a^2c^2d^2+6a^2cd^3+2ab^3c^4+4ab^3c^3d+2ab^3c^2d^2+6ab^2c^4+8ab^2c^3d+4ab^2c^2d^2+4ab^2cd^3+6abc^4+4abc^3d+6abc^2d^2+12abcd^3+6abd^4+2ac^4+4ac^2d^2+6acd^3+2b^3c^3d+2b^3c^2d^2+4b^2c^3d+4b^2c^2d^2+2b^2cd^3+2bc^3d+4bc^2d^2+4bcd^3+2bd^4+2c^2d^2+2cd^3$$ denominator: $$a^3b^3c^4+4a^3b^3c^3d+6a^3b^3c^2d^2+4a^3b^3cd^3+a^3b^3d^4+3a^3b^2c^4+10a^3b^2c^3d+12a^3b^2c^2d^2+6a^3b^2cd^3+a^3b^2d^4+3a^3bc^4+8a^3bc^3d+7a^3bc^2d^2+2a^3bcd^3+a^3c^4+2a^3c^3d+a^3c^2d^2+a^2b^3c^4+6a^2b^3c^3d+12a^2b^3c^2d^2+10a^2b^3cd^3+3a^2b^3d^4+3a^2b^2c^4+15a^2b^2c^3d+24a^2b^2c^2d^2+15a^2b^2cd^3+3a^2b^2d^4+3a^2bc^4+12a^2bc^3d+14a^2bc^2d^2+5a^2bcd^3+a^2c^4+3a^2c^3d+2a^2c^2d^2+2ab^3c^3d+7ab^3c^2d^2+8ab^3cd^3+3ab^3d^4+5ab^2c^3d+14ab^2c^2d^2+12ab^2cd^3+3ab^2d^4+4abc^3d+8abc^2d^2+4abcd^3+ac^3d+ac^2d^2+b^3c^2d^2+2b^3cd^3+b^3d^4+2b^2c^2d^2+3b^2cd^3+b^2d^4+bc^2d^2+bcd^3$$ status: 0 From weighted AM-GM inequality: $$2a^2b^2c^3d \le a^3bc^2d^2+ab^3c^4$$ $$2a^2c^3d \le a^3c^2d^2+ac^4$$ $$4a^2bc^3d \le a^3bc^2d^2+a^3c^2d^2+ab^3c^4+ac^4$$ $$ 0 \le 4a^3bcd^3+2a^3bd^4+2a^3cd^3+2a^2b^2cd^3+4a^2bc^2d^2+12a^2bcd^3+6a^2bd^4+4a^2c^2d^2+6a^2cd^3+4ab^3c^3d+2ab^3c^2d^2+6ab^2c^4+8ab^2c^3d+4ab^2c^2d^2+4ab^2cd^3+6abc^4+4abc^3d+6abc^2d^2+12abcd^3+6abd^4+4ac^2d^2+6acd^3+2b^3c^3d+2b^3c^2d^2+4b^2c^3d+4b^2c^2d^2+2b^2cd^3+2bc^3d+4bc^2d^2+4bcd^3+2bd^4+2c^2d^2+2cd^3 $$ The sum of all inequalities gives us a proof of the inequality.
0