2021/041) The equation $(x+a)(x+b)=9$ has a root a+b . Prove that $ab \le 1$

 Solution 

Because a+b is a root so $(x+a)(x+b)=9$ shall be satisfied when x = a + b.
Putting x = a + b we get
$(2a+b)(2b+a) = 9$
or $2a^2 + 2b^ 2 + 5ab = 9$
or $2(a^2+b^2) + 5ab = 9$
or $2(a^2+b^2-2ab) + 9ab = 9$
or $2(a- b)^2 = 9(1-ab)$
now LHS is >=0 so is RHS so $9(1-ab) \ge 0$ or $1-ab \ge 0$ or $ab \le 1$