TITLE: Prove $\sqrt{x^2-x+1}\ge\log{\left(\frac{e^x+e}{2}\right)}$
QUESTION [6 upvotes]: The question: Let $x\in \Bbb{R}$,  show that
$$\sqrt{x^2-x+1}\ge\log{\left(\dfrac{e^x+e}{2}\right)}$$
My attempt:
I have proven that 
$$\sqrt{x^2-x+1}\ge\dfrac{1}{2}(x+1)$$
since the right-hand side is the tangent line to $f(x)=\sqrt{x^2-x+1}$ at $x=1$.
So it suffices to prove
$$\dfrac{1}{2}(x+1)\ge \log{\left(\dfrac{e^x+e}{2}\right)},$$
but I haven't managed to do this.

REPLY [9 votes]: Set $t=e^{x-1}$. Need to prove:
$$\sqrt{\mathrm{log}(t)(\mathrm{log}(t)+1)+1}\ge \mathrm{log}(\frac{et+e}{2})$$
$$\Leftrightarrow f(t)=\mathrm{log}^2(t)+\mathrm{log}(t)-\mathrm{log}^2(\frac{t+1}{2})-2\mathrm{log}(\frac{t+1}{2})\ge0$$
$$\Leftarrow f'(t)\ge0,t\ge1;f'(t)\le0,t\le1$$
$$\Leftrightarrow g(t)\ge0,t\ge1;g(t)\le0,t\le1$$
where $g(t)=(t+1)\mathrm{log}(t)-t\mathrm{log}(\frac{t+1}{2})-\frac{t-1}{2}, t\in(0,+\infty)$.
Now we would like to prove $g(t)$ to be increasing over $(0,+\infty)$.
$\displaystyle{g'(t)=\mathrm{log}(t)+\frac{t+1}{t}-\mathrm{log}(\frac{t+1}{2})-\frac{t}{t+1}-\frac{1}{2}\\
\qquad =\mathrm{log}(1+\frac{t-1}{t+1})+\frac{t+1}{t}-\frac{t}{t+1}-\frac{1}{2}\\
\qquad\ge\frac{t-1}{2t}+\frac{t+1}{t}-\frac{t}{t+1}-\frac{1}{2}\\
\qquad=\frac{3t+1}{2t(t+1)}>0}$
Q.E.D.