I am trying to prove the inequality $$\ln x\leq\frac{x^{\frac{x+\sqrt{x}+1}{2\sqrt{x}}}-1}{\sqrt{x}}$$ is true for all positive real numbers, without using calculus.
I realized the equality occurs if and only if for $x=1$.
My progress. Let $\sqrt{x}=y$ we obtain $$\ln y^2\leq\frac{y^{\frac {y^2+y+1}{y}}-1}{y}$$ or $$\ln y\leq\frac{y^{\frac{y^2+y+1}{y}}-1}{2y}$$
I have no idea how to proceed further.
Note that the inequality is false below $x=1$. For $y\ge 1$, $1.5(y-1) \ge \ln(y)$ (there are various ways to see this, taking a derivative is one of the ways). But once you are convinced of the above, $$\ln(y)\le \frac{3(y-1)}{2}\le \frac{(y-1)(y+1+1/y)}{2} \le \frac{y^3-1}{2y}\le \frac{y^{y+1+1/y}-1}{2y} = \frac{y^{\frac{y^2+y+1}{y}} - 1}{2y}.$$ I only use the inequality $y+1/y \ge 2$ for this.
Let $r(x)=\frac{x^{\frac{x+\sqrt{x}+1}{2 \sqrt{x}}}-1}{\sqrt{x}}$. Then the inequality $$\ln x\leq r(x)$$ is false for $x<1$ since $r(1)=\ln(1)=0$ but $r^{'}(1)=1.5$ and $\ln(x)$ has derivative $1$ at $x=1$. Alternatively you can just check that the inequality fails for $x=0.5$ since $r(0.5)<-.9<-.7<\ln(0.5)$.
However it is true for all $x\geq 1$ with equality iff $x=1$.
Proof
If $x\geq 1$ then $\sqrt x\geq1$ and hence $x^{3/2}-\sqrt x\leq x^{3/2}-1$ and $x-1\leq \frac{x^{3/2}-1}{\sqrt x}$.
By AM-GM, for $x>0$, $\frac{x+\sqrt{x}+1}{2 \sqrt{x}}\geq 3/2$ with equality iff $x=1$. Since $x\geq1$ this implies $x^{3/2}\leq x^{\frac{x+\sqrt{x}+1}{2 \sqrt{x}}}$ and therefore $x-1\leq \frac{x^{3/2}-1}{\sqrt x}\leq r(x)$ with equality iff $x=1$. But $\ln x \leq x-1$ for $x>0$ with equality iff $x=1$. Hence $\ln x\leq r(x)$ for $x\geq 1$ with equality iff $x=1$.
