# Prove that $\ln x\leq\frac{x^{\frac{x+\sqrt{x}+1}{2\sqrt{x}}}-1}{\sqrt{x}}$ for all positive real numbers

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.

• Use derivatives or rather calculus Commented Aug 21, 2023 at 11:21
• Use the inequality $y+1/y \ge 2$ repeatedly Commented Aug 21, 2023 at 12:52

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 $$x=1$$.
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 $$x=1$$.