Prove that the inequality holds for all real numbers a and b.Prove that the inequality: (a+5b)(3a+2b) greater than or equal to (a+9b)(2a+b)

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You need to prove that `(a+5b)(3a+2b) >= (a+9b)(2a+b)` , hence, you should open the brackets such that:

`3a^2 + 2ab + 15ab + 10b^2 >= 2a^2 + ab + 18ab + 9b^2`

You need to move all terms to the left and you need to combine similar terms such that:

`(3a^2 - 2a^2) + (17ab - 19ab) + (10b^2 - 9b^2) >= 0`

`a^2 - 2ab + b^2 >= 0`

Notice that the expression `a^2 - 2ab + b^2`  is the development of the binomial (a-b)^2 such that:

`(a-b)^2 >= 0`

You should know that the square of a number is always positive, hence, the last inequality proves that the initial inequality `(a+5b)(3a+2b) >= (a+9b)(2a+b)`  holds for all `a, b in R.`