# Please prove the inequalityProve that the inequality (a+5b)(3a+2b)≥(a+9b)(2a+b) Holds real for all real numbers a and b.

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### 2 Answers

You need to open the brackets both sides such that:

`(a+5b)(3a+2b)≥(a+9b)(2a+b)`

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

Moving all terms to one side yields:

`3a^2 + 2ab + 15ab + 10b^2- 2a^2- ab- 18ab- 9b^2 >= 0`

Collecting like terms yields:

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

Notice that if you expand the square `(a-b)^2` yields `a^2 - 2ab + b^2` , hence, you may substitute `(a-b)^2 ` for `a^2 - 2ab + b^2` such that:

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

**Notice that a square will always be positive, regardless the values of a and b and the square is zero if the values of a and b are equal, hence, the inequality `(a+5b)(3a+2b)≥(a+9b)(2a+b)` holds for all real a and b.**

(a+5b)(3a+2b)≥(a+9b)(2a+b) , multiply out

3a^2 + 2ab + 15ab + 10b^2 ≥ 2a^2 + ab + 18ab + 9b^2 , add 'ab' terms together

3a^2 + 17ab + 10b^2 ≥ 2a^2 + 19ab + 9b^2 , algebraically move all variables to left side of inequality

a^2 - 2ab + b^2 ≥ 0 , factor

(a-b)(a-b) ≥ 0 , same as

(a-b)^2 ≥ 0 , any real number squared is positive or 0 (i.e. ≥ 0)