How is $(\log x)^4 \leq x^3$ $\forall x>1$

Question

Here is my trying: $$\forall x > 1 [\log(x) \leq x]$$ Taking 4th power both sides we get. $$\forall x > 1 [(\log(x))^4 \leq x^4]$$

But in the book they used the inequality: $$\forall x > 1 [(\log(x))^4 \leq x^3]$$

How can we prove this?

Answer 1

Let $f(x)=x^{\frac{3}{4}}-\ln{x}$.

Hence, $f'(x)=\frac{3}{4\sqrt[4]x}-\frac{1}{x},$ which gives $x_{min}=\frac{4}{3}\sqrt[3]{\frac{4}{3}}$

and since $f\left(x_{min}\right)=0.949...>0$, we are done!

Answer 2

Let $x=t^4$; the inequality becomes $$ (\log(t^4))^4\le t^{12} $$ which is equivalent, for $t\ge1$, to $$ 4\log t\le t^3 $$ Oh, well, we can also set $t=\sqrt[3]{u}$, so the inequality becomes $$ \frac{4}{3}\log u\le u $$ Consider $f(u)=3u-4\log u$, defined over $[1,\infty)$. Then $f(1)=3$ and $$ f'(u)=3-\frac{4}{u}=\frac{3u-4}{u} $$ showing $f$ has a minimum at $u=4/3$; since $$ f(4/3)=4-4\log(4/3)>0 $$ (because $4/3<e$), we are done.

Answer 3

Let's prove the strict inequality $\log x\lt x^{3/4}$ for $x\gt1$ without resorting to derivatives, using instead the inequalities $2\lt e^{3/4}$ and $k+1\le2^k$ for all $k\in\mathbb{N}$.

Each $x\gt1$ can be written (uniquely) in the form $x=e^ku$ with $k\in\mathbb{N}$ and $1\lt u\le e$. Since $\log u\le1$ for such $u$, we have

$$\log x=k+\log u\le k+1\le2^k\le e^{3k/4}\lt(e^ku)^{3/4}=x^{3/4}$$

(The strict inequality is due to $1\lt u$. The non-strict inequalities allow for $u=e$, $k\le1$, and $k=0$, respectively.)

For completeness, the inductive proof that $k+1\le2^k$ starts with the base case $0+1=2^0$ and then, using $k\ge0$, finds that $(k+1)+1\le2(k+1)\le2\cdot2^k=2^{k+1}$. As for the inequality $2\lt e^{3/4}$, one proof is to establish the equivalent inequality $\log2\lt{3\over4}$, which can be done by drawing a picture of the geometric interpretation of the integral definition $\log2=\int_1^2{dt\over t}$, noting that the area beneath the curve lies entirely within a trapezoid of area ${1\over2}(1+{1\over2})$. Alternatively, if we start from $2.7\lt e$, then

$$16\cdot10^3=(16\cdot5)\cdot5^2\cdot2^3=80\cdot25\cdot8\lt3^4\cdot3^3\cdot3^2=3^9=27^3$$

implies $2^4=16\lt(27/10)^3\lt e^3$. If you want a proof that $2.7\lt e$, we have, from the Taylor series for $e$,

$$e\gt1+1+{1\over2}+{1\over6}+{1\over24}=2+{12+4+1\over24}=2+{17\over24}$$

and ${17\over24}\gt{7\over10}$ since $17\cdot5=85\gt84=7\cdot12$.

Answer 4

Start with:

$e^{3y} \ge y^4$ for $y \gt 0$.

$e^{3y} \ge \frac{ (3y)^4}{4!} = \frac{3^4}{4!} y^4 \ge y^4$.

Recall the series expansion of $e^{3y}$:

$e^{3y} = 1 + (3y) + \frac{(3y)^2}{2!} + \frac{(3y)^3}{3!} + \frac{(3y)^4}{4!} .....$

Let $y = \log x$, for $x \gt 1$

$e^{3\log x} \ge (\log x)^4$;

$x^3 \ge (\log x)^4$ for $x \ge 1$.

Note: $e^{3\log x} = e^{\log (x^3)} = x^3$.