Find the difference between the maximum and minimum values of $a+b$
that satisfy
$$a+b+\frac{1}{a}+\frac{9}{b}=10,\quad(a,b\in\mathbb{R}^+)$$
I'm trying to use Cauchy–Schwarz inequality, but I can't get a sense of how to transform and solve the given equation. Can anyone help me?
Best Answer
Using the CS-inequality "Engels" form we have: $10 = a+b+\dfrac{1^2}{a}+\dfrac{3^2}{b} \ge a+b+\dfrac{(1+3)^2}{a+b}=a+b+\dfrac{16}{a+b}=x+\dfrac{16}{x}\implies 10x\ge x^2+16\implies x^2-10x+16 \le 0\implies (x-2)(x-8) \le 0\implies 2 \le x = a+b \le 8 \implies \text{max - min} = 8 - 2 = 6$. The minimum value of $a+b$ is $2$ which occurs when $a+b=2, \dfrac{1}{a} = \dfrac{3}{b} \implies a = \dfrac{1}{2}, b = \dfrac{3}{2}$. The maximum of $a+b$ is $8$ which occurs when $a+b = 8, b = 3a \implies a=2,b=6$.