How to prove #(a+5b)(a+2b)=9b(a+b)# given a, b, c, d, are any 4 positive numbers?

Question

How to prove #(a+5b)(a+2b)>=9b(a+b)# given a, b, c, d, are any 4 positive numbers?

Julia Adams · Accepted Answer

See a solution process below:

We can expand both sides of the equation as: #(a * a) + (a * 2b) + (5b * a) + (5b * 2b) >= (9b * a) + (9b * b)# #a^2 + 2ab + 5ab + 10b^2 >= 9ab + 9b^2# #a^2 + 7ab + 10b^2 >= 9ab + 9b^2# We can now subtract #color(red)(9ab)# and #color(blue)(9b^2)# from each side of the equation to put all of the variables on one side of the equation while keeping the equation balanced: #a^2 + 7ab - color(red)(9ab) + 10b^2 - color(blue)(9b^2) >= 9ab - color(red)(9ab) + 9b^2 - color(blue)(9b^2)# #a^2 + (7 - color(red)(9))ab + (10 - color(blue)(9))b^2 >= 0 + 0# #a^2 + (-2)ab + 1b^2 >= 0 + 0# #a^2 - 2ab + b^2 >= 0# We can now factor this as: #(a - b)(a - b) >= 0# #(a - b)^2 >= 0# Any number squared, by definition, will be a greater than or equal to zero. If #(a - b) < 0# - then a minus times a minus is a plus and greater than zero. If #(a - b) > 0# - then a plus times a plus is a plus and greater than zero. If #(a - b) = 0# - then a zero times zero is zero and equal to zero.

How to prove #(a+5b)(a+2b)&gt;=9b(a+b)# given a, b, c, d, are any 4 positive numbers?

How to prove #(a+5b)(a+2b)>=9b(a+b)# given a, b, c, d, are any 4 positive numbers?