An inequality involving 2 variables

Question

How to prove that the following expression is positive for $u\ge 0$ and $q\in(0,\pi/3)$. $$\frac{2 \sqrt{2} ((2+u) (1+\cos(q)))^{3/4}}{3^{3/4}}-\frac{2^{3/4} (1+\sec(q))}{\sqrt{3} \sec(q)^{3/4}}-\left(1+u+\sqrt{-1+(1+u)^2} \sqrt{1-\frac{4 \sin(q)^2}{3}}\right)^{3/4}$$

Answer 1

Letting $c:=\cos q$, we have $c\in[1/2,1]$. Writing $\sec q=1/c$ and $\sin^2q=1-c^2$, and multiplying the big expression by $3^{3/4}c^{1/4}$, we see that the inequality in question can be rewritten as \begin{equation*} f(c,u)\overset{\text{(?)}}>0, \tag{10}\label{10} \end{equation*} where $(c,u)\in[1/2,1]\times[0,\infty)$, \begin{equation*} f(c,u):=\big(1-(r_1(c,u)+r_2(c,u))\big)\,2\sqrt2\, (c (1 + c)^3 (2 + u)^3)^{1/4}, \end{equation*} \begin{equation*} r_1(c,u):=\frac{\sqrt[4]{3}}{2^{3/4}}\, \sqrt[4]{\frac{c+1}{c (u+2)^3}},\quad r_2(c,u):=\frac1{2 \sqrt{2}}\, \left(\frac{\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 u+3}{c u+2 c+u+2}\right)^{3/4}. \end{equation*}

So, it is enough to show that \begin{equation*} r_1(c,u)+r_2(c,u)\overset{\text{(?)}}<1. \tag{20}\label{20} \end{equation*}

Note that $r_1(c,u)$ is decreasing in $c\in[1/2,1]$ and in $[0,\infty)$. So, \begin{equation*} r_1(c,u)\le r_1(1/2,73/10)<1-\frac{3^{3/4}}{2 \sqrt{2}}\quad \text{for }(c,u)\in[1/2,1]\times[73/10,\infty). \end{equation*} Next, for all $(c,u)\in[1/2,1]\times[0,\infty)$ \begin{equation*} \begin{aligned} &\big((2\sqrt2\,r_2(c,u))^{4/3}-3\big) \\ &\times (c+1) (u+2) \left(\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 c u+6 c+3\right) \\ &=\left(3-3 c^2\right) u^2+6 \left(2 c^2+3 c+1\right) u+9 (2 c+1)^2, \end{aligned} \end{equation*} which is manifestly $>0$, whence $r_2(c,u)<\dfrac{3^{3/4}}{2 \sqrt{2}}$. So, we have \eqref{20} and hence \eqref{10} for $(c,u)\in[1/2,1]\times[73/10,\infty)$.

It remains to prove \eqref{10} for $(c,u)\in[1/2,1]\times[0,73/10]$.

To do this, note that \begin{equation} f(c,u)=f_1(c,u)-f_2(c,u), \end{equation} where \begin{equation} f_1(c,u):=2 \sqrt{2} \sqrt[4]{c (c+1)^3 (u+2)^3}-2^{3/4} \sqrt[4]{3} (c+1), \end{equation} \begin{equation} f_2(c,u):=\sqrt[4]{c \left(\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 u+3\right)^3}. \end{equation} Clearly, $f_2(c,u)$ is increasing in $c\in[1/2,1]$ and in $[0,\infty)$.

Also clearly, $f_1(c,u)$ is increasing in $u\in[0,\infty)$. Also, \begin{equation} g_1(c,u):=\partial_c f_1(c,u)=\frac1{\sqrt{2}} \Big(\frac{(4 c+1) (u+2)^{3/4}}{\sqrt[4]{c^3 (c+1)}}-2 \sqrt[4]{6}\Big) \ge g_1(c,0) \end{equation} and \begin{equation} \partial_c g_1(c,0)=-\frac{3}{2\ 2^{3/4} c^{7/4} (c+1)^{5/4}}<0. \end{equation} So, $g_1(c,u)\ge g_1(c,0)\ge g_1(1,0)=2.78\ldots>0$. So, $f_1(c,u)$ is increasing in $c\in[1/2,1]$. So, $f_1(c,u)$ is increasing in $c\in[1/2,1]$ and in $[0,\infty)$.

So, "partitioning" the rectangle $[0,73/10]$ into $20\times200=4000$ congruent rectangles, with vertices $(c_i,u_j):=(\frac{1}{2}+\frac{1}{2}\frac{i}{20},\frac{73}{10}\frac{j}{200})$, for all $(c,u)\in[1/2,1]\times[0,73/10]$ we get \begin{equation} \begin{aligned} f(c,u) \ge\min\big\{f_1 & \big(c_{i-1},u_{j-1}\big) -f_2\big(c_i,u_j\big) \colon \\ &i=1,\dots,20,\,j=1,\dots,200\big\}=0.0247\ldots>0. \end{aligned} \end{equation} This completes the proof of \eqref{10}. $\quad\Box$

Answer 2

First let's multiply the given expression by $\left(\frac92\sec(q)\right)^{3/4}$ and simplify it into $$\left(6(2+u) (\sec(q)+1)\right)^{3/4}-3 (1+\sec(q))-\left(\frac92(1+u)\sec(q)+\frac92\sqrt{-1+(1+u)^2} \sqrt{\frac{4-\sec(q)^2}{3}}\right)^{3/4}$$ Let $v:=\sec(q)$ and thus $v\in[1,2]$. Then, let $p:=(6(2+u) (v+1))^{1/4}$ and $r:=\left(\frac92(1+u)v+\frac92\sqrt{u^2+2u} \sqrt{\frac{4-v^2}{3}}\right)^{1/4}$. Correspondingly, the original question is equivalent to showing that the following system of polynomial equations and inequalities is insoluble: $$\begin{cases} p^4 - 6(2+u) (v+1) = 0, \\ (2r^4 - 9(1+u)v)^2 - 27(u^2+2u)(4-v^2) = 0,\\ p\geq0, \\ r\geq0, \\ u\geq 0, \\ 1 \leq v \leq 2, \\ p^3 - 3(1+v) - r^3 \leq 0. \end{cases}$$ This formulation makes the problem amenable to existing software. A few examples and links can be seen in the answers to this question.

Answer 3

Remarks: Here is a human verifiable proof. The same strategy as my answer for another question of the same author is used. The proof of (1), (2) and (3) is not difficult by using Bernoulli inequality and letting $x = \cos q \in [1/2, 1]$, and thus omitted here.

For convenience, let $$a := \frac{2 \sqrt{2} (1+\cos q)^{3/4}}{3^{3/4}}, \quad b := \frac{2^{3/4} (1+\sec q)}{\sqrt{3}\, (\sec q)^{3/4}}, \quad c := \sqrt{1-\frac{4 \sin^2 q}{3}}.$$

We need to prove that, for all $u \ge 0$ and $q \in [0, \pi/3]$, $$a(2 + u)^{3/4} - b + \left(1 + u + c\sqrt{-1 + (1 + u)^2}\right)^{3/4} \ge 0.$$

We split into three cases.

Case 1: $u > 10/9$

Using $\sqrt{-1 + (1 + u)^2} < 1 + u$, it suffices to prove that $$g(u) := a(2 + u)^{3/4} - b - (1 + c)^{3/4}(1 + u)^{3/4} \ge 0.$$

We have \begin{align*} g'(u) &= \frac{3a}{4(2 + u)^{1/4}} - \frac{3(1 + c)^{3/4}}{4(1 + u)^{1/4}}\\[6pt] &= \frac{3(1 + c)^{3/4}}{4(2 + u)^{1/4}} \left(\frac{a}{(1 + c)^{3/4}} - \left(1 + \frac{1}{1 + u}\right)^{1/4}\right)\\[6pt] &\ge \frac{3(1 + c)^{3/4}}{4(2 + u)^{1/4}} \left(\frac{a}{(1 + c)^{3/4}} - \left(1 + \frac{1}{1 + 10/9}\right)^{1/4}\right)\\[6pt] &> 0 \end{align*} where we use the fact that, for all $q \in [0, \pi/3]$, $$\left(\frac{a}{(1 + c)^{3/4}}\right)^{4/3} - \left(1 + \frac{1}{1 + 10/9}\right)^{1/3} = \frac{4(1 + \cos q)}{3\left(1 + \sqrt{1-\frac{4 - 4 \cos^2 q}{3}}\right)} - \sqrt[3]{\frac{28}{19}} > 0.$$ (Note: Simply let $x = \cos q \in [1/2, 1]$. The rest is easy.)

Also, we have $$g(10/9) = a(2 + 10/9)^{3/4} - b - (1 + c)^{3/4}(1 + 10/9)^{3/4} > 0. \tag{1}$$

Thus, we have $g(u) > 0$ for all $u > 10/9$.

$\phantom{2}$

Case 2: $1/5 < u \le 10/9$

We have $\sqrt{-1 + (1 + u)^2} = 1 + u - \frac{1}{1 + u + \sqrt{-1 + (1 + u)^2}} \le 1 + u - \frac14$.

It suffices to prove that $$h(u) := a(2 + u)^{3/4} - b + \left(1 + u + c(u + 3/4)\right)^{3/4} \ge 0.$$

Letting $d := \frac{1 + 3c/4}{1 + c}$, we have \begin{align*} h'(u) &= \frac{3a}{4(2+u)^{1/4}} - \frac{3(1 + c)^{3/4}}{4(u + d)^{1/4}}\\[6pt] &= \frac{3(1 + c)^{3/4}}{4(2 + u)^{1/4}}\left(\frac{a}{(1 + c)^{3/4}} - \left(1 + \frac{2 - d}{u + d}\right)^{1/4}\right)\\[6pt] &\ge \frac{3(1 + c)^{3/4}}{4(2 + u)^{1/4}}\left(\frac{a}{(1 + c)^{3/4}} - \left(16/7\right)^{1/4}\right)\\[6pt] & > 0 \end{align*} where we use $1 + \frac{2-d}{u + d} \le 1 + \frac{2-d}{d} = \frac{8 + 8c}{4 + 3c} \le 16/7$ (using $c\le 1$), and the fact that, for all $q \in [0, \pi/3]$, $$\left(\frac{a}{(1 + c)^{3/4}}\right)^{4/3} - \left(16/7\right)^{1/3} = \frac{4(1 + \cos q)}{3\left(1 + \sqrt{1-\frac{4 - 4 \cos^2 q}{3}}\right)} - (16/7)^{1/3} > 0.$$ (Note: Simply let $x = \cos q \in [1/2, 1]$. The rest is easy.)

Also, we have $$h(1/5) = a(2 + 1/5)^{3/4} - b + \left(1 + 1/5 + c(1/5 + 3/4)\right)^{3/4} > 0. \tag{2}$$

Thus, we have $h(u) > 0$ for all $1/5 < u \le 10/9$.

$\phantom{2}$

Case 3: $0\le u \le 1/5$

By Bernoulli inequality $(1 + z)^r \le 1 + zr$ for all $0 < r < 1$ and $z > -1$, we have $$(2 + u)^{3/4} = \frac{2 + u}{2^{1/4} (1 + \frac{u}{2})^{1/4}} \ge \frac{2 + u}{2^{1/4}(1 + \frac{u}{2}\cdot \frac14)} \ge 2^{3/4}(1 + u/3)$$ where we use $2 + u \ge 2(1 + u/3)(1 + u/8)$.

Also, we have $\sqrt{-1 + (1 + u)^2} \le \sqrt{-1 + (1 + 1/5)^2} < 2/3$.

It suffices to prove that $$F(u) := a\cdot 2^{3/4}(1 + u/3) - b + \left(1 + u + 2c/3\right)^{3/4} \ge 0.$$

We have \begin{align*} F'(u) &= \frac{2^{3/4}}{3}a - \frac{3}{4(1 + u + 2c/3)^{1/4}} \\[6pt] &\ge \frac{2^{3/4}}{3}\cdot \frac{2 \sqrt{2} (1+1/2)^{3/4}}{3^{3/4}} - \frac{3}{4}\\ &> 0 \end{align*} where we use $a \ge \frac{2 \sqrt{2} (1+1/2)^{3/4}}{3^{3/4}}$.

Also, we have $$F(0) = a\cdot 2^{3/4} - b + \left(1 + 2c/3\right)^{3/4} > 0. \tag{3}$$

Thus, we have $F(u) > 0$ for all $u \in [0, 1/5]$.

We are done.