If $x > y$, can you prove $x \log y > y \log x$, $x \ge 1$ and $y \ge 1$

Question

If $x > y$, can you prove $x \ \log y > y \log x$, where $x \ge 1$ and $y \ge 1$. I encountered this problem in a paper I read and somehow cannot prove it.

In questions like this, separate the variables - x on one side, y on the other - and show that the resulting function is monotonic - usually by differentiating. — A.S., Mar 12 '16 at 22:54

score 1 · Accepted Answer · answered Mar 12 '16 at 22:57

Consider the function $$ f(x)=\frac{\log x}{x} $$ defined for $x\ge1$. Note that $f(1)=0$. Also $$ f'(x)=\frac{1-\log x}{x^2} $$ So the function has a maximum at $e$ and the claim is false. If you assume $e\le y<x$, then, as $f$ is decreasing on $[e,\infty)$, you have $$ f(y)>f(x) $$ that is, $$ \frac{\log y}{y}>\frac{\log x}{x} $$ that becomes $$ x\log y>y\log x $$

Note that, instead, if $1\le y<x\le e$, you have $$ f(y)<f(x) $$ because $f$ is increasing on $[1,e]$. This translates to $$ x\log y<y\log x $$

score 0 · Answer 2 · answered Mar 12 '16 at 22:56

Define

$$f(x)=\frac{\log x}x\implies f'(x)=\frac{1-\log x}{x^2}>0\iff \log x<1\iff 0<x<e$$

Thus, for $\;x>e\;$ we get that $\;f'(x)<0\implies f(x)\;$ monotonic descending and then

$$\forall\,x>y>e\;,\;\;f(x)<f(y)\iff y\log x<x\log y$$

Thus, not for all $\;x,y\ge1\;$ .

If $x > y$, can you prove $x \log y > y \log x$, $x \ge 1$ and $y \ge 1$

2 Answers2