Recall the Cauchy-Schwartz inequality: for all x, y Element R^n, |x middot y| lessthanorequalto ||x|| || y || Prove that the square of the average of a list of numbers is less than or equal to the average of the squares of the numbers. That is, show that for all X_1, ..., x_n element R, (1/n sigma^n_i = 1 x_i)^2 lessthanorequalto 1/n sigma &n_i = 1 x_i^2. Without using calculus, prove that for all theta Element R, cos(theta) + sin(theta) lessthanorequalto Squareroot 2.