How to prove $x_1+x_2\in (2,3)$ if $x_1

Question

I have tried many methods but failed. How to prove this?

I can see that '3',for $x_1\in (0,1),x_2\in (\sqrt2,2)$. Using matlab I got the sum is about 2.2, greater than 2. But how to get this '2'?

Answer 1

Let $$f(x) = (3/4)^x - |\log_2 x|.$$

We have $f(1/2) < 0$, $f(1) = 3/4$, $f(3/2) > 0$, and $f(2) < 0$.

For $0 < x < 1$, we have $f(x) = (3/4)^x + \log_2 x$ and $$f'(x) = (3/4)^x \ln \frac34 + \frac{1}{x\ln 2} > \ln \frac34 + \frac{1}{\ln 2} > 0.$$ Thus, $f(x) = 0$ has exactly one real solution (say $x_1$) on $(0, 1)$; furthermore, $1/2 < x_1 < 1$.

For $x > 1$, we have $f(x) = (3/4)^x - \log_2 x$ and $$f'(x) = (3/4)^x\ln \frac34 - \frac{1}{x\ln 2} < 0.$$ Thus, $f(x) = 0$ has exactly one real solution (say $x_2$) on $[1, \infty)$; furthermore, $3/2 < x_2 < 2$.

Thus, we have $2 < x_1 + x_2 < 3$.

We are done.