Consider for any positive number $a$, how to show that $\sqrt{x} + \sqrt{x} \sqrt{\ln{x}} \leq 2 + 2x$?

Question

Consider for any positive number $x$, how to show that $\sqrt{x} + \sqrt{x} \sqrt{\ln{x}} \leq 2 + 2x$?

I tried Cauchy inequality and Taylor expansion but still can't figure it out.

Answer 1

Recalling that for $x \ge 1$, $$\ln x < x$$ and $$\sqrt{x}\le x$$ you have $$\sqrt{x} + \sqrt{x} \sqrt{\ln x} < \sqrt{x} + \sqrt{x} \sqrt{x} = \sqrt{x} + x \le x+x = 2x < 2x+2$$

For $x < 1$, $\sqrt{ \ln x}$ is not defined.