An inequality for a real function

Question

Let $$f(z)=(1+z)^{3/4}-\left(\frac{3}{8}+\frac{\sqrt{3}}{4}\right)^{1/4}-\frac{\left(3 z+\sqrt{6} \sqrt{-1+z^2}\right)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}}.$$ Is there a simple proof that this function is positive for $z\ge 1$?

Answer 1

Make the substitution $u=(1+z)^{3/4}$, so that $u\ge2^{3/4}$, $z=u^{4/3}-1$, and the inequality in question becomes $$F(u):=f(u^{4/3}-1)>0. \tag{1}\label{1}$$

For $z=u^{4/3}-1>1$ $$F''(u)=\frac{5 u^2 \sqrt{z-1}+\sqrt{6} \left(2 z^2+3 z-1\right)}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4} (z-1)^{3/2} (z+1) \left(\sqrt{6} \sqrt{z^2-1}+3 z\right)^{5/4}}, $$ which is manifestly $>0$. So, $F(u)$ is convex in $u\ge2^{3/4}$. Also, for $u_*:=\frac{212}{100}$ we have $F'(u_*)=0.00227\ldots>0$. So, $$\min_{u\ge2^{3/4}}F(u) \ge\min_{u\ge2^{3/4}}[F(u_*)+F'(u_*)(u-u_*)] \\ =F(u_*)+F'(u_*)(2^{3/4}-u_*)=0.0579\ldots>0.$$ Thus, \eqref{1} is proved. $\quad\Box$

Answer 2

Alternative proof.

We split into two cases.

Case 1: $z > 2$

Using $3 z+\sqrt{6} \sqrt{-1+z^2} \le 3 z+\sqrt{6}\, z = (3 + \sqrt 6)z$, it suffices to prove that $$g(z) := (1+z)^{3/4}-\left(\frac{3}{8}+\frac{\sqrt{3}}{4}\right)^{1/4}-\frac{\left(3+\sqrt{6}\right)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}}\cdot z^{3/4} > 0.$$

We have \begin{align*} g'(z) &= \frac{3}{4(1 + z)^{1/4}} \left(1 - \left(1 + \frac{1}{z}\right)^{1/4}\frac{(3 + \sqrt 6)^{3/4}}{(4 + 2\sqrt 3)^{3/4}}\right)\\[6pt] &\ge \frac{3}{4(1 + z)^{1/4}} \left(1 - \left(1 + \frac{1}{2}\right)^{1/4}\frac{(3 + \sqrt 6)^{3/4}}{(4 + 2\sqrt 3)^{3/4}}\right)\\[6pt] &> 0. \end{align*} Also, $g(2) > 0$. Thus, $g(z) > 0$ for all $z > 2$.

$\phantom{2}$

Case 2: $1 \le z \le 2$

Using $z - \sqrt{z^2-1} = \frac{1}{z + \sqrt{z^2-1}} \ge \frac{1}{2 + \sqrt{3}}$, we have $3z + \sqrt{6}\sqrt{-1+z^2} \le (3 + \sqrt 6)z - \frac{\sqrt 6}{2 + \sqrt 3}$.

It suffices to prove that \begin{align*} h(z) &:= (1+z)^{3/4}-\left(\frac{3}{8}+\frac{\sqrt{3}}{4}\right)^{1/4} -\frac{(3 + \sqrt 6)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}} \cdot \left(z -a\right)^{3/4} > 0 \end{align*} where $a := \frac{\sqrt 6}{(2 + \sqrt 3)(3 + \sqrt 6)}$.

We have \begin{align*} h'(z) &= \frac{3}{4(z - a)^{1/4}} \left(\left(1 - \frac{1 + a}{1 + z}\right)^{1/4} - \frac{(3 + \sqrt 6)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}}\right)\\[6pt] &\ge \frac{3}{4(z - a)^{1/4}} \left(\left(1 - \frac{1 + a}{1 + 1}\right)^{1/4} - \frac{(3 + \sqrt 6)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}}\right)\\[6pt] &> 0. \end{align*} Also, $h(1) > 0$. Thus, $h(z) > 0$ for all $z \in [1, 2]$.

We are done.