Inequality of an expression greater than 0

Question

Wolfram says the following is true for all $x>0$:

$$-(x+1)^{(x+1) }+ 2^{(x+1)} x^x\geq0$$

How to prove this algebraically?

Answer 1

Since $x >0$, instead of considering function $$f(x)= 2^{(x+1)} x^x-(x+1)^{(x+1) }$$ consider instead $$g(x)=\log\left(2^{(x+1)} x^x\right) -\log\left((x+1)^{(x+1) } \right)$$ $$g(x)=(x+1)\log(2)+x \log(x)-(x+1)\log(x+1)$$ Now, the derivatives $$g'(x)=\log (x)-\log (x+1)+\log (2)\qquad \text{and} \qquad g''(x)=\frac{1}{x^2+x}\qquad > 0 \qquad \forall x >0$$ The first derivative cancels at $x=1$. For this value $g(1)=0$ and by the second serivative test, this is a minimum. So, no root for $g(x)=0$ which is a decreasing function for $ 0 < x <1$, goes through a minimum and increases again (for ever).