If m is an odd integer, prove that (m4 + 4m2 + 11) / 16 is an integer. If a, b, c are positive real numbers, prove that sqrt(b/a) sqrt(c/b) sqrt(a/c) = 1.

Step 1: Solve Question 1. Let m be an odd integer. We need to show that the number (m^4 + 4m^2 + 11)/(16) is an integer. Since m is an odd integer, we can write m = 2k+1 for some integer k. First, let's find m^2: m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1 The product of two consecutive integers, k(k+1), is always an even number. So, k(k+1) = 2j for some integer j. Therefore, m^2 = 4(2j) + 1 = 8j + 1. This means m^2 1 8. Next, let's find m^4: m^4 = (m^2)^2 = (8j+1)^2 = 64j^2 + 16j + 1 = 16(4j^2 + j) + 1 This means m^4 1 16. Let p = 4j^2 + j, which is an integer. So, m^4 = 16p + 1. Now substitute m^4 = 16p+1 and m^2 = 8j+1 into the given expression: (m^4 + 4m^2 + 11)/(16) = ((16p+1) + 4(8j+1) + 11)/(16) = (16p + 1 + 32j + 4 + 11)/(16) = (16p + 32j + 16)/(16) Factor out 16 from the numerator: = (16(p + 2j + 1))/(16) = p + 2j + 1 Since p and j are integers, their sum p+2j+1 is also an integer. Thus, the number (m^4 + 4m^2 + 11)/(16) is an integer. The expression simplifies to an integer p+2j+1. Step 2: Solve Question 2. Given that a, b, c are positive real numbers, we need to show that sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1. We can use the property of square roots that sqrt(x) × sqrt(y) = sqrt(xy) and sqrt((x)/(y)) = sqrt(x)sqrt(y). Combine the terms under a single square root: sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = sqrt((b)/(a) × (c)/(b) × (a)/(c)) Multiply the fractions inside the square root: = sqrt((b · c · a)/(a · b · c)) Cancel out the common terms in the numerator and denominator: = sqrt(1) = 1 Thus, sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1. sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1 Step 3: Solve Question 3. Given sqrt(n) = (3)/(2). We need to find -sqrt(n). Substitute the given value of sqrt(n) into the expression -sqrt(n): -sqrt(n) = -((3)/(2)) -sqrt(n) = -(3)/(2) -(3)/(2) Step 4: Solve Question 4. Let a and b be positive real numbers. We need to show that (a)/(b) + (b)/(a) 2. We know that for any real number x, (x-y)^2 0. Consider the expression (sqrt((a)/(b)) - sqrt((b)/(a)))^2. Since a and b are positive, sqrt((a)/(b)) and sqrt((b)/(a)) are real numbers. Therefore, (sqrt((a)/(b)) - sqrt((b)/(a)))^2 0 Expand the square: (sqrt((a)/(b)))^2 - 2sqrt((a)/(b))sqrt((b)/(a)) + (sqrt((b)/(a)))^2 0 (a)/(b) - 2sqrt((a)/(b) · (b)/(a)) + (b)/(a) 0 (a)/(b) - 2sqrt(1) + (b)/(a) 0 (a)/(b) - 2 + (b)/(a) 0 Add 2 to both sides of the inequality: (a)/(b) + (b)/(a) 2 This proves the statement. (a)/(b) + (b)/(a) 2 Step 5: Solve Question 5. Let x be a real number. We need to show that x^2 + (1)/(x^2) 2, for x 0. We know that the square of any real number is non-negative. Consider the expression (x - (1)/(x))^2. Since x is a real number and x 0, x - (1)/(x) is a real number. Therefore, (x - (1)/(x))^2 0 Expand the square: x^2 - 2 · x · (1)/(x) + ((1)/(x))^2 0 x^2 - 2 + (1)/(x^2) 0 Add 2 to both sides of the inequality: x^2 + (1)/(x^2) 2 This proves the statement. x^2 + (1)/(x^2) 2