I have problem with following inequality $\log_{4}{5}+\log_{5}{6}+\log_{6}{7}+\log_{7}{8} \ge 4.4$
I tried to change all logarithms to base $10$ but it didn't work
[Math] logarithmic inequality with different bases
logarithms
Related Solutions
$$2-\log_3(x-7)=\log_{\frac13}(2x)$$ $$2-\frac{\log(x-7)}{\log3}=\frac{\log(2x)}{\log{\frac13}}$$ $$2-\frac{\log(x-7)}{\log3}=\frac{\log(2x)}{-\log{3}}$$ $$2\log(3)-\log(x-7)=-\log(2x)$$ $$\log{\frac{9}{x-7}}=\log{\frac{1}{2x}}$$ $$\frac{9}{x-7}=\frac{1}{2x}$$ $$18x=x-7$$ $$x=\frac{-7}{17}$$ But for logarithm to be defined, we must have $x-7\gt0$ and $2x\gt0$. Thus, there is no solution.
We want to solve $$ \ x \log_{\log_{f(x)} (g(x)) } \left( h(x)\right) \geq 0$$
where $$f(x)=|x^2 - 3 | - 2,\quad g(x)=x^2 - 3|x| + 2,\quad h(x)= \dfrac{x^3 - |3x+2|}{x^3 - |3x-2|}$$
First of all, $x=0$ is not a solution since $f(0)=1$.
Comparing $x$ with $0$, $f(x)$ with $1$, $g(x)$ with $f(x)$, $h(x)$ with $1$, we have eight cases to consider :
Case 1 : $x\lt 0$ and $0\lt f(x)\lt 1$ and $f(x)\lt g(x)\lt 1$ and $h(x)\ge 1$
Case 2 : $x\lt 0$ and $f(x)\gt 1$ and $1\lt g(x)\lt f(x)$ and $h(x)\ge 1$
Case 3 : $x\lt 0$ and $0\lt f(x)\lt 1$ and $0\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$
Case 4 : $x\lt 0$ and $f(x)\gt 1$ and $g(x)\gt f(x)$ and $0\lt h(x)\le 1$
Case 5 : $x\gt 0$ and $0\lt f(x)\lt 1$ and $f(x)\lt g(x)\lt 1$ and $0\lt h(x)\le 1$
Case 6 : $x\gt 0$ and $f(x)\gt 1$ and $1\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$
Case 7 : $x\gt 0$ and $0\lt f(x)\lt 1$ and $0\lt g(x)\lt f(x)$ and $h(x)\ge 1$
Case 8 : $x\gt 0$ and $f(x)\gt 1$ and $g(x)\gt f(x)$ and $h(x)\ge 1$
Using the following lemmas :
Lemma 1 : If $x\lt 0$, then $h(x)\lt 1$.
Lemma 2 : There are no $x$ such that $f(x)\gt 1$ and $g(x)\gt f(x)$.
(The proofs for the lemmas are written at the end of the answer.)
we see that Case 1, Case 2, Case 4, Case 8 don't happen.
Now, we have four cases to consider :
Case 3 : $x\lt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $0\lt h(x)\le 1$
Case 5 : $x\gt 0$ and $0\lt f(x)\lt g(x)\lt 1$ and $0\lt h(x)\le 1$
Case 6 : $x\gt 0$ and $1\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$
Case 7 : $x\gt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $h(x)\ge 1$
Case 3 : $x\lt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $0\lt h(x)\le 1$
Since $x\lt 0$, we have $g(x)=x^2-3(-x)+2=x^2+3x+2$.
Case 3-1 : For $x\in(-\infty,-\sqrt 3)$, we have $$f(x)=(x^2-3)-2=x^2-5,\quad h(x)=\frac{x^3-(-(3x+2))}{x^3-(-(3x-2))}=\frac{x^3+3x+2}{x^3+3x-2}$$ $$0\lt g(x)\lt f(x)\lt 1\iff 0\lt x^2+3x+2\lt x^2-5\lt 1\iff x\in\left(-\sqrt 6,-\frac 73\right)$$ Since $x^3+3x-2\lt 0$ for $x\lt 0$, $$0\lt h(x)\le 1\iff 0\lt \frac{x^3+3x+2}{x^3+3x-2}\le 1\iff x^3+3x-2\le x^3+3x+2\lt 0$$ $$\iff x^3+3x+2\lt 0\iff x\lt \alpha$$ where $\left(-\frac 73\lt\right)\alpha$ is the only root of $i(x)=x^3+3x+2$ which is increasing with $i(-7/3)\lt 0$.
So, in this case, we have $$x\in\left(-\sqrt 6,-\frac 73\right)\tag1$$
Case 3-2 : For $x\in\left[-\sqrt 3,-\frac 23\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(-(3x+2))}{x^3-(-(3x-2))}=\frac{x^3+3x+2}{x^3+3x-2}$$
$$0\lt g(x)\lt f(x)\lt 1\iff 0\lt x^2+3x+2\lt -x^2+1\lt 1\iff x\in\left(-1,-\frac 12\right)$$
$$0\lt h(x)\le 1\iff x\lt \alpha$$where $\left(-\frac 23\lt\right)\alpha$ is the only root of $i(x)=x^3+3x+2$ which is increasing with $i(-2/3)\lt 0$.
So, in this case, we have $$x\in\left(-1,-\frac 23\right)\tag2$$
Case 3-3 : For $x\in\left[-\frac 23,0\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(-(3x-2))}=\frac{x^3-3x-2}{x^3+3x-2}$$
$$0\lt g(x)\lt f(x)\lt 1\iff x\in\left(-1,-\frac 12\right)$$ $$0\lt h(x)\le 1\iff 0\lt \frac{x^3-3x-2}{x^3+3x-2}\le 1\iff x^3+3x-2\le x^3-3x-2\lt 0$$ $$\iff x^3-3x-2=(x-2)(x+1)^2\lt 0\iff x\in(-\infty,-1)\cup (-1,2)$$ So, in this case, we have $$x\in \left[-\frac 23,-\frac 12\right)\tag3$$
Therefore, in Case 3, we have $$(1)\cup (2)\cup (3)\iff x\in\left(-\sqrt 6,-\frac 73\right)\cup \left(-1,-\frac 12\right)\tag4$$
Case 5 : $x\gt 0$ and $0\lt f(x)\lt g(x)\lt 1$ and $0\lt h(x)\le 1$
Since $x\gt 0 $, we have $g(x)=x^2-3x+2$.
Case 5-1 : For $x\in\left(0,\frac 23\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(-(3x-2))}=\frac{x^3-3x-2}{x^3+3x-2}$$ $$0\lt f(x)\lt g(x)\lt 1\iff 0\lt -x^2+1\lt x^2-3x+2\lt 1\iff x\in\left(\frac{3-\sqrt 5}{2},\frac 12\right)$$ Since $x^3+3x-2\lt 0$ for $x\lt \frac 12$ where $j(x)=x^3+3x-2$ is increasing with $j(1/2)\lt 0$, $$0\lt h(x)\le 1\iff 0\lt \frac{x^3-3x-2}{x^3+3x-2}\le 1\iff x^3+3x-2\le x^3-3x-2\lt 0$$ which does not hold for $x\gt 0$.
Case 5-2 : For $x\in\left[\frac 23,\sqrt 3\right)$, we have $f(x)=-x^2+1$ and $$0\lt f(x)\lt g(x)\lt 1\iff x\in\left(\frac{3-\sqrt 5}{2},\frac 12\right)$$ There are no $x$ such that $x\in\left[\frac 23,\sqrt 3\right)\cap \left(\frac{3-\sqrt 5}{2},\frac 12\right)$.
Case 5-3 : For $x\in [\sqrt 3,\infty)$, we have $$f(x)=(x^2-3)-2=x^2-5,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(3x-2)}=\frac{x^3-3x-2}{x^3-3x+2}$$ $$0\lt f(x)\lt g(x)\lt 1\iff 0\lt x^2-5\lt x^2-3x+2\lt 1\iff x\in\left(\sqrt 5,\frac 73\right)$$ Since $x^3-3x+2=(x+2)(x-1)^2\gt 0$ for $x\ge \sqrt 3$, $$0\lt h(x)\le 1\iff 0\lt \frac{x^3-3x-2}{x^3-3x+2}\le 1\iff 0\lt x^3-3x-2\le x^3-3x+2$$ $$\iff 0\lt x^3-3x-2=(x-2)(x+1)^2\iff x\in (2,\infty)$$ Therefore, in Case 5, we have $$x\in\left(\sqrt 5,\frac 73\right)\tag5$$
Case 6 : $x\gt 0$ and $1\lt g(x)\lt f(x)$ and $0\lt h(x)\le 1$
Since $x\gt 0$, we have $g(x)=x^2-3x+2$.
Case 6-1 : For $x\in\left(0,\sqrt 3\right)$, we have $f(x)=-(x^2-3)-2=-x^2+1$ and $$1\lt g(x)\lt f(x)\iff 1\lt x^2-3x+2\lt -x^2+1$$ There are no such $x$.
Case 6-2 : For $x\in [\sqrt 3,\infty)$, we have $$f(x)=(x^2-3)-2=x^2-5,\quad h(x)=\frac{x^3-3x-2}{x^3-3x+2}$$
$$1\lt g(x)\lt f(x)\iff 1\lt x^2-3x+2\lt x^2-5\iff x\in\left(\frac{3+\sqrt 5}{2},\infty\right)$$ $$0\lt h(x)\le 1\iff x\in (2,\infty)$$
Therefore, in Case 6, we have $$x\in\left(\frac{3+\sqrt 5}{2},\infty\right)\tag6$$
Case 7 : $x\gt 0$ and $0\lt g(x)\lt f(x)\lt 1$ and $h(x)\ge 1$
Since $x\gt 0$, we have $g(x)=x^2-3x+2$.
Case 7-1 : For $x\in\left(0,\frac 23\right)$, we have $$f(x)=-(x^2-3)-2=-x^2+1,\quad h(x)=\frac{x^3-(3x+2)}{x^3-(-(3x-2))}=\frac{x^3-3x-2}{x^3+3x-2}$$ $$0\lt g(x)\lt f(x)\lt 1\iff 0\lt x^2-3x+2\lt -x^2+1\lt 1\iff x\in\left(\frac 12,1\right)$$ Since $j(x)=x^3+3x-2$ is increasing with $j(\beta)=0$ where $\frac 12\lt \beta\lt \frac 23$, we have
For $x\lt\beta$ where $x^3+3x-2\lt 0$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\le x^3+3x-2\iff x\ge 0$$
For $x\gt\beta$ where $x^3+3x-2\gt 0$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\ge x^3+3x-2\iff x\le 0$$
from which we have $$h(x)\ge 1\iff 0\lt x\lt\beta$$
So, in this case, we have $$x\in\left(\frac 12,\beta\right)$$ where $\beta\approx 0.596$ is the only real root of $x^3+3x-2$.
Case 7-2 : For $x\in \left[\frac 23,\infty\right)$, we have $$h(x)=\frac{x^3-(3x+2)}{x^3-(3x-2)}=\frac{x^3-3x-2}{x^3-3x+2}$$ Since $x^3-3x+2=(x+2)(x-1)^2\gt 0$ for $x\in \left[\frac 23,\infty\right)$ with $x\not=1$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3-3x+2}\ge 1\iff x^3-3x-2\ge x^3-3x+2$$ There are no such $x$.
Therefore, in Case 7, we have $$x\in\left(\frac 12,\beta\right)\tag7$$ where $\beta\approx 0.596$ is the only real root of $x^3+3x-2$.
Hence, the answer is $(4)\cup (5)\cup (6)\cup (7)$, i.e. $$\small\color{red}{x\in\left(-\sqrt 6,-\frac 73\right)\cup \left(-1,-\frac 12\right)\cup \left(\frac 12,\sqrt[3]{1+\sqrt 2}-\frac{1}{\sqrt[3]{1+\sqrt 2}}\right)\cup \left(\sqrt 5,\frac 73\right)\cup \left(\frac{3+\sqrt 5}{2},\infty\right)}$$ where $\sqrt[3]{1+\sqrt 2}-\frac{1}{\sqrt[3]{1+\sqrt 2}}=\beta\approx 0.596$ is the only real root of $x^3+3x-2$.
(From Sid's comment, we can find $\beta$ by setting $x=y-\frac 1y$ for $x^3+3x-2=0$ to have $(y^3)^2-2y^3-1=0$ which is a quadratic equation on $y^3$.)
Finally, let us prove Lemma 1 and Lemma 2.
Lemma 1 : If $x\lt 0$, then $h(x)\lt 1$.
Proof :
$$h(x)\ge 1\iff \frac{x^3-|3x+2|}{x^3-|3x-2|}\ge 1$$
For $x\in\left(-\infty,-\frac 23\right)$, since $x^3+3x-2\lt 0$, $$h(x)\ge 1\iff \frac{x^3+3x+2}{x^3+3x-2}\ge 1\iff x^3+3x+2\le x^3+3x-2$$which does not hold.
For $x\in\left[-\frac 23,\beta\right)$ where $\beta$ is the only real root of $x^3+3x-2$, since $x^3+3x-2\lt 0$, $$h(x)\ge 1\iff\frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\le x^3+3x-2\iff x\ge 0$$
For $x\in\left(\beta,\frac 23\right)$, since $x^3+3x-2\gt 0$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3+3x-2}\ge 1\iff x^3-3x-2\ge x^3+3x-2\iff x\le 0$$
For $x\in\left[\frac 23,\infty\right)$, since $x^3-3x+2=(x+2)(x-1)^2\gt 0$ with $x\not=1$, $$h(x)\ge 1\iff \frac{x^3-3x-2}{x^3-3x+2}\ge 1\iff x^3-3x-2\ge x^3-3x+2$$which does not hold.
So, the claim follows from that $h(x)\ge 1\implies x\ge 0$. $\quad\blacksquare$
Lemma 2 : There are no $x$ such that $f(x)\gt 1$ and $g(x)\gt f(x)$.
Proof :
$$f(x)\gt 1\iff |x^2-3|-2\gt 1\iff |x^2-3|\gt 3$$ $$\iff x^2-3\lt -3\quad\text{or}\quad x^2-3\gt 3\iff x\in (-\infty,-\sqrt 6)\cup (\sqrt 6,\infty)$$ On the other hand, $$g(x)\gt f(x)\iff x^2-3|x|+2\gt |x^2-3|-2$$
For $x\in (-\infty,-\sqrt 6)$, $$g(x)\gt f(x)\iff x^2-3(-x)+2\gt (x^2-3)-2\iff x\in \left(-\frac 73,\infty\right)$$
For $x\in (\sqrt 6,\infty)$, $$g(x)\gt f(x)\iff x^2-3x+2\gt (x^2-3)-2\iff x\in\left(-\infty,\frac 73\right)$$
The claim follows from that $\frac 73\lt \sqrt 6$. $\quad\blacksquare$
Best Answer
Rewrite the inequality as
$${1\over4}\left({\log5\over\log4}+{\log6\over\log5}+{\log7\over\log6}+{\log8\over\log7}\right)\gt{11\over10}$$
Now apply the AM-GM inequality:
$${1\over4}\left({\log5\over\log4}+{\log6\over\log5}+{\log7\over\log6}+{\log8\over\log7}\right)\ge\sqrt[4]{\log8\over\log4}=\sqrt[4]{3\over2}$$
It remains to observe that
$${3\over2}\gt\left({11\over10}\right)^4={14641\over10000}$$