Need help with a tricky inequality problem

Question

I need help with this inequality:

Given that $y \geq 1$ and $x > 0$ and knowing that $y \geq 4 x \ln(x)$ show that this implies $y \geq x \ln(2y)$.

I saw this problem while reading a theoretical machine learning paper and it says this inequality follows but I have attempted many things including going directly from the assumption to working backwards. Seeing on desmos the inequality does seem to hold for the values of x,y listed. Also it may be important to note that y is a sample size (so natural numbers). Thank you!

Answer 1

You are given that $y \ge 1$, $x \gt 0$ and

$$y \ge 4x\ln(x) \tag{1}\label{eq1A}$$

Then it's asked to show this implies

$$y \ge x\ln(2y) \tag{2}\label{eq2A}$$

This can be done by splitting the range of $x$ into $2$ cases:

Case $1$: $x \le 1$

We have $\ln(x) \le 0 \implies 4x\ln(x) \le 0$, so \eqref{eq1A} is true since $y \ge 1$. Next, set

$$f(y) = y - \ln(2y) \tag{3}\label{eq3A}$$

This gives $f(1) = 1 - \ln(2) \approx 0.307 \gt 0$. Also, since $y \ge 1$, then

$$f'(y) = 1 - \frac{1}{y} \ge 0 \tag{4}\label{eq4A}$$

This shows $f(y)$ is an increasing function, and thus always positive, so

$$y - \ln(2y) \gt 0 \implies y \gt \ln(2y) \ge x\ln(2y) \tag{5}\label{eq5A}$$

which proves that \eqref{eq2A} holds.

Case $2$: $x \gt 1$

Define

$$g(x) = y - x\ln(2y) \tag{6}\label{eq6A}$$

Since $y \ge 1 \implies \ln(2y) \gt 0$, then for $x \gt 0$, note $g(x)$ is a strictly decreasing function in $x$, with it reaching $0$ at

$$x_1 = \frac{y}{\ln(2y)} \tag{7}\label{eq7A}$$

so for any $x \lt x_1$ we have $g(x) \gt 0$. Next, define

$$h(y; x) = y - 4x\ln(x) \tag{8}\label{eq8A}$$

Since both $4x$ and $\ln(x)$ are positive and strictly increasing functions, then $4x\ln(x)$ is strictly increasing in $x$, so $-4x\ln(x)$ is strictly decreasing. Thus, if $h(y; x_1) \lt 0$, then there's an $x_2 \lt x_1$ such that $h(y; x_2) = 0$. As such, only $x \le x_2$ satisfies \eqref{eq1A}, with these $x$ also satisfying \eqref{eq2A}. To show this, first note

$$\begin{equation}\begin{aligned} h(y; x_1) & = y - \frac{4y\ln\left(\frac{y}{\ln(2y)}\right)}{\ln(2y)} \\ & = y\left(1 - 4\left(\frac{\ln(y) - \ln(\ln(2y))}{\ln(2y)}\right)\right) \end{aligned}\end{equation}\tag{9}\label{eq9A}$$

Next, since $y \ge 1$, then the sign of $h(y; x_1)$ is the same as that of

$$h_1(y) = 1 - 4\left(\frac{\ln(y) - \ln(\ln(2y))}{\ln(2y)}\right) \tag{10}\label{eq10A}$$

Using the quotient rule and chain rule when differentiating gives

$$\begin{equation}\begin{aligned} h'_1(y) & = -4\left(\frac{\left(\frac{1}{y} - \frac{1}{\ln(2y)}\left(\frac{1}{y}\right)\right)\ln(2y) - \left(\ln(y) - \ln(\ln(2y)\right)\left(\frac{1}{y}\right)}{(\ln(2y))^2}\right) \\ & = \left(\frac{-4}{y(\ln(2y))^2}\right)\left(\ln(2y) - 1 - \ln(y) + \ln(\ln(2y))\right) \\ & = \left(\frac{-4}{y(\ln(2y))^2}\right)\left(\ln(2) - 1 + \ln(\ln(2y))\right) \end{aligned}\end{equation}\tag{11}\label{eq11A}$$

This gives

$$h'_2(y_1) = 0 \implies \ln(\ln(2y_1)) = 1 - \ln(2) \tag{12}\label{eq12A}$$

Note $y_1 \approx 1.946$, with $h'_1(y) \gt 0$ for $y \lt y_1$ and $h'_1(y) \lt 0$ for $y \gt y_1$. Thus, $h_1(y_1) \approx -0.05704$ is a maximum (with the corresponding $h(y_1; x_1) \approx -0.111$). Since this is $\lt 0$, then as discussed previously, this means that if \eqref{eq1A} holds, then so does \eqref{eq2A} in this case as well.

Answer 2

If $x^4>2y$ finish because $y> x ln(x^4)$, how $x^4> 2y$ then $y> x ln(x^4)>xln(2y)$.

Now just prove $x^4>2y$.

$y> 4xln(x)$ then when you calculate $x^4 > 2(4xln(x))$ you will see $x>0$ is the solution.

For x=0 this inequality don't exist

Answer 3

Using elementary calculus, it is easy to see that $f:[1,\infty)\to[0,\infty), f(t)=t\ln(t)$ is (strictly) increasing and 1:1 correspondence.

As $\frac{y}{4}\ge\frac{1}{4}>0$, there exists $t>1$ with $t\ln(t)=\frac{y}{4}$

Substituting $y$ for $t$ in the given inequality one gets $t\ln(t)\ge x\ln(x)$, which (under the condition $t>1$) is equivalent to $t\ge x$.

Then the inequality to be proven can be rewritten as:

$$4t\ln(t)\ge x\ln(8t\ln(t))$$

As $t\ge x>0$, it suffices to prove that $4\ln(t)\ge\ln(8t\ln(t))$, or equivalently:

$$t^3\ge8\ln(t)$$

This last inequality is trivial to prove with elementary calculus. Note that in fact the strict inequality ($y>x\ln(2)$) holds.