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?
A number \(n\) is an integer such that \(n\) is not divisible by \(3\) or by \(2\). Show that \(n^2-1\) is divisible by \(24\).
Show that for any two positive real numbers \(x,y\) it is true that \(x^2+y^2 \ge 2xy\).