For a natural number \(n\) prove that \(n! \leq (\frac{n+1}{2})^n\), where \(n!\) is the factorial \(1\times 2\times 3...\times n\).
Prove the \(AM-GM\) inequality for \(n=2\). Namely for two non-negative real numbers \(a,b\) we have \(2\sqrt{ab} \leq a+b\).
Prove the Cauchy-Schwartz inequality: for a natural number \(n\) and real numbers \(a_1, a_2, ... a_n\) and \(b_1, b_2, ...b_n\) we have \[(a_1b_1 + a_2b_2 + ...a_nb_n)^2 \leq (a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...b_n^2).\]
Prove the \(HM-GM\) inequality for positive real numbers \(a_1,a_2,...a_n\): \[\frac{n}{\frac{1}{a_1} + ... \frac{1}{a_n}} \leq \sqrt[n]{a_1a_2...a_n}.\]
From 1999 IMO. Let \(n\geq 2\) be an integer. Determine the least possible constant \(C\) such that the inequality: \[\sum_{1\leq i<j\leq n} x_ix_j(x_i^2 + x_j^2) \leq C(\sum_{1\leq i\leq n}x_i)^4\] holds for all non-negative real numbers \(x_i\). For this constant \(C\) find out when the equality holds.
Find all pairs of whole numbers \((x,y)\) so that this equation is true: \(xy+1 = y+x\).
Albert was calculating consecutive squares of natural numbers and looking at differences between them. He noticed the difference between \(1\) and \(4=2^2\) is \(3\), the difference between \(4\) and \(9=3^2\) is \(5\), the difference between \(9\) and \(16=4^2\) is \(7\), between \(16\) and \(5^2=25\) is \(9\), between \(25\) and \(6^2=36\) is \(11\). Find out what the rule is and prove it.
Find all pairs of whole numbers \((x,y)\) so that this equation is true: \(xy = y+1\).
After Albert discovered the previous rule, he began looking at differences of squares of consecutive odd numbers. He found the difference between \(1^2\) and \(3^2\) is \(8\), the difference between \(3^2\) and \(5^2\) is \(16\), the difference between \(5^2\) and \(7^2\) is \(24\), and that the difference between \(7^2\) and \(9^2\) is \(32\). What is the rule now? Can you prove it?