Solve Let \(a\), \(b\), and \(c\) be positive real numbers such that \(a + b + c = 3\).Prove the following inequality:\(a3b+c+b3c+a+c3a+b (3)/(2)\)

Step 1: Rewrite the denominators using the given condition. Given a+b+c=3, we can express the denominators as: b+c = 3-a c+a = 3-b a+b = 3-c Substituting these into the inequality, we get: a^33-a+b^33-b+c^33-c (3)/(2) Step 2: Define a function and check its convexity. Let f(x) = (x^3)/(3-x) for x (0, 3). Since a, b, c are positive and sum to 3, each must be less than 3. We need to find the second derivative of f(x) to check for convexity. First, find the first derivative f'(x): f'(x) = (d)/(dx)((x^3)/(3-x)) = (3x^2(3-x) - x^3(-1))/((3-x)^2) = (9x^2 - 3x^3 + x^3)/((3-x)^2) = (9x^2 - 2x^3)/((3-x)^2) Next, find the second derivative f''(x): f''(x) = (d)/(dx)((9x^2 - 2x^3)/((3-x)^2)) Using the quotient rule ((u)/(v))' = (u'v - uv')/(v^2): Let u = 9x^2 - 2x^3 u' = 18x - 6x^2 Let v = (3-x)^2 v' = 2(3-x)(-1) = -2(3-x) f''(x) = ((18x - 6x^2)(3-x)^2 - (9x^2 - 2x^3)(-2(3-x)))/((3-x)^4) Factor out (3-x) from the numerator: f''(x) = ((3-x)[(18x - 6x^2)(3-x) + 2(9x^2 - 2x^3)])/((3-x)^4) f''(x) = ((18x - 6x^2)(3-x) + 2(9x^2 - 2x^3))/((3-x)^3) Expand the numerator: f''(x) = ((54x - 18x^2 - 18x^2 + 6x^3) + (18x^2 - 4x^3))/((3-x)^3) f''(x) = (2x^3 - 18x^2 + 54x)/((3-x)^3) Factor out 2x from the numerator: f''(x) = (2x(x^2 - 9x + 27))/((3-x)^3) Step 3: Analyze the sign of the second derivative. For x (0, 3): • 2x > 0 • (3-x)^3 > 0 • For the quadratic x^2 - 9x + 27, its discriminant is = (-9)^2 - 4(1)(27) = 81 - 108 = -27. Since < 0 and the leading coefficient (1) is positive, x^2 - 9x + 27 > 0 for all real x. Therefore, f''(x) > 0 for all x (0, 3). This means f(x) is a convex function on the interval (0, 3). Step 4: Apply Jensen's Inequality. Since f(x) is convex, by Jensen's inequality: (f(a)+f(b)+f(c))/(3) f((a+b+c)/(3)) Substitute f(x) = (x^3)/(3-x) and the given condition a+b+c=3: (1)/(3)(a^33-a+b^33-b+c^33-c) ((a+b+c)/(3))^33-((a+b+c)/(3)) (1)/(3)(a^33-a+b^33-b+c^33-c) ((3)/(3))^33-((3)/(3)) (1)/(3)(a^33-a+b^33-b+c^33-c) (1^3)/(3-1) (1)/(3)(a^33-a+b^33-b+c^33-c) (1)/(2) Step 5: Multiply by 3 to obtain the desired inequality. a^3b+c+b^3c+a+c^3a+b (3)/(2) The inequality is proven. That's 2 down. 3 left today — send the next one.