Algebraic inequalities

Can tackle this problem as follows:

.

it is impossible that a & b are real numbers where (a²+1)(b²+4)=10ab-5.

(a^2+1)(b^2+4) = 10ab - 5     

WolframAlpha shows that  there are two solutions :

x =  √(3/2) , y = √6

And 

x = - √(3/2),  y = -√6

We only need to check one of these........

( 3/2 + 1) (6 + 4)  = 10(√(3/2)(√6) - 5

(5/2) (10)  = 10√(18/2) - 5

25  = 10√9  - 5

25  = 10*3 - 5

25 = 30 - 5

25 = 25

You can verify that the other solution will check, as well

So the value of a^2 + b^2 =  3/2 + 6 =   3/2 + 12/2  =  15/2  =  7.5