How to show $2-2/x - \log_2(x) \geq 0$ for $x \in [1,2]$?

Question

I have an answer to Binary tree where every internal vertex is of degree 3., but it requires a proof of the following:

Question: How to show $2-2/x - \log_2(x) \geq 0$ for $x \in [1,2]$?

Wolfram|Alpha indicates it's true, although you need to click on "approximate form" and "more digits", when it says it's true when 1.000000000000000000<=x<=2.0000000000000000000.

We have equality when $x=1$ and $x=2$, so the left-hand side is not increasing on the interval $[1,2]$, which makes it hard to prove via derivatives.

Answer 1

Hint: define $$f(x)=2-\frac{2}{x}-\frac{\ln(x)}{\ln(2)}$$ and calculate $$f(1),f(2)$$ and solve the equation $$f'(x)=0$$ for $x$

Answer 2

Let $$f(x)=2-\frac2x-\frac{\log x}{\log2}$$ Then $$\int_1^2 f(x)\,dx=\left[2x-2\log x-\frac{x(\log x-1)}{\log 2}\right]_1^2=2-2\log2-\frac{2(\log2-1)}{\log2}-\frac1{\log2}>0$$ Now all you have to do is show that $x=1,2$ are the only points where $f$ crosses the $x$-axis.

Answer 3

The function $$ f(x) = 2 - \frac 2x - \frac{\ln x}{\ln 2} $$ is (strictly) concave in the interval $[1, 2]$ since $$ f''(x) = -\frac{4}{x^3} + \frac{1}{x^2 \ln 2 } = \frac{1}{x^2 \ln 2} \left( 1- \frac {4 \ln 2}x \right) \le \frac{ 1- 2 \ln 2 }{x^2 \ln 2} < 0 $$ for $1 \le x \le 2$, therefore its graph lies above the line segment from $(1, f(1))$ to $(2, f(2))$.

You already noticed that $f(1) = f(2) = 0$, so the conclusion follows.

(More generally, any concave function on a closed interval attains it minimum at one of the interval end points.)