Show that 4 rectangles of dimensions a * b can fit in a square with side length a + b.
+506 The area of 4 rectangles of dimensions a*b has an area of 4ab, while a square with side length a+b has an area of (a+b)^2=a^2+2ab+b^2. We want to show:
\(4ab\le a^2+2ab+b^2\\0\le a^2-2ab+b^2\\ 0 \le (a-b)^2\)
Since \((a-b)^2\) will always be nonnegative for real numbers a and b, it will always be greater or equal to zero. \(\square\)
+506 The area of 4 rectangles of dimensions a*b has an area of 4ab, while a square with side length a+b has an area of (a+b)^2=a^2+2ab+b^2. We want to show:
\(4ab\le a^2+2ab+b^2\\0\le a^2-2ab+b^2\\ 0 \le (a-b)^2\)
Since \((a-b)^2\) will always be nonnegative for real numbers a and b, it will always be greater or equal to zero. \(\square\)
