Solutions to review problems

This page is going to contain some of the solutions I submit to mathematical periodicals with problem sections such as Mathematical Reflections and The American Mathematical Monthly. Problems are usually rewritten in order to avoid excessive quoting and often to homogenize the notations used in a problem and its solution.

I don't put up solutions on this site prior to the discussion of the problem in the respective magazine, so usually they won't appear here directly after the deadline is over.

Mathematical Reflections

• Solution to: Titu Andreescu, Problem U111, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the second published solution) as a PDF file.
Let n be a positive integer. For every k in {0, 1, ..., n-1}, let ak = 2 cos(pi / 2n-k). Prove that product_{k=0}^{n-1} (1-ak) = (-1)n-1 / (1 + a0).

• Solution to: Cezar Lupu and Valentin Vornicu, Problem U112, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the first published solution) as a PDF file.
Let x, y, z be real numbers greater or equal to 1. Prove that x^{x³+2xyz} y^{y³+2xyz} z^{z³+2xyz} >= (xxyyzz)yz+zx+xy.

• Solution to: Titu Andreescu, Problem O111, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the second published solution) as a PDF file.
Prove that, for each integer n >= 0, the number (binom(n,0) + 2 binom(n,2) + 22 binom(n,4) + ...)² (binom(n,1) + 2 binom(n,3) + 22 binom(n,5) + ...)² is triangular.
Here, binom(n,m) means the (n,m)-th binomial coefficient (that is, n(n-1)...(n-m+1) / m! if m >= 0, and 0 otherwise).

• Solution to: Cezar Lupu and Pham Huu Duc, Problem O112, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (not published) as a PDF file.
Let a, b, c be positive real numbers. Prove that
(a³+abc) / (b+c)² + (b³+abc) / (c+a)² + (c³+abc) / (a+b)² >= 3/2 * (a³+b³+c³)/(a²+b²+c²).

• Solution to: Gabriel Dospinescu, Problem O114, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the first published solution) as a PDF file.
Prove that for all real numbers x, y, z, the following inequality holds:
(y²+yz+z²) (z²+zx+x²) (x²+xy+y²) >= 3(x²y+y²z+z²x) (xy²+yz²+zx²).

The American Mathematical Monthly

Currently solved (and solutions submitted): #11391, #11392, #11393, #11395, #11397, #11398, #11401, #11402, #11403, #11406, #11407, #11409, #11417.
I am planning to put up solutions to #11391, #11397, #11403, #11406, and #11407, as well as all other solutions that don't make it into the journal here when time comes.

Project PEN (Problems in Elementary Number Theory)

• Solution to: Problem E16, a. k. a.: Mathematics Magazine, Problem 1392 by George Andrews.
(This links to my solution on the PEN server. A local version can be found here: Solution to Project PEN Problem E 16.)

If n is a positive integer, and p is a prime lying in the interval ]n, 4n/3], then prove that p divides sum_{j=0}^{n} binom(n,j) 4, where binom(n,j) means n! / (j! (n-j)!).

My solution (which generalizes the problem three times) is an edited and extended version of my posting in MathLinks topic #150539 (which only generalizes it one time).

Solutions to review problems

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Darij Grinberg