Show that for arbitrary a, b, u > 0 one has
(a + u) 2 a + b · u ≥ a + b−2 b a 2 a + b b−2 b a = 4ab − 1 b2 .
Hint: Show that the function
f(u) = (a + u) 2 a + b · u
satisfies
f (u) <><>
and
f(u) > 0 if u > b − 2 b · a.
Already registered? Login
Not Account? Sign up
Enter your email address to reset your password
Back to Login? Click here