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Imagine an inequality of the form:

$$\frac{f(x + y)}{y(y^2-1)} < \frac{f(x) + f(y)}{y(y^2-1)}$$

Then, is it true that

$$ \displaystyle \int \int_{[a,b]\times[a,b]} \frac{f(x + y)}{y(y^2-1)} dy dx < \int \int_{[a,b]\times[a,b]} \left( \frac{( f(x) + f(y)}{y(y^2-1)} \right) dy dx$$ ?

And that

$$ \displaystyle \int_a^b \frac{f(x + y)}{y(y^2-1)} dy < \int_a^b \left( \frac{f(x) + f(y)}{y(y^2-1)} \right) dy$$ ?

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    $\begingroup$ The first question doesn't quite make sense. The limit are just $a$ and $b$, but you write $dydx$ as if it is a 2-dimensional integral. Do you mean it to be over $[a,b]\times[a,b]$ or something? $\endgroup$ – Thompson Feb 5 '17 at 16:08
  • $\begingroup$ Yes, sorry. Corrected $\endgroup$ – user3141592 Feb 5 '17 at 16:28
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Set $g_1(y)= \frac{f(x+y)}{y(y^2-1)}$ and $g_2(y)= \frac{f(x)+f(y)}{y(y^2-1)}$. As $g_1 < g_2$ you have your second assertion. Then you consider the integration against $x$ and you have your first assertion. No?

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  • $\begingroup$ Then, is it as easy as multiplying both sides by $dy$? And aren't there any limitations on $a$ and $b$? $\endgroup$ – user3141592 Feb 5 '17 at 16:06
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    $\begingroup$ Yes, I supposed $a<b$. But if you are not in this case the first assertion is false. $\endgroup$ – Babyblog Feb 5 '17 at 16:14
  • $\begingroup$ My real question: and why can't yo apply that to $x<y$ to make it $\int_a^b x dy < \int_a^b y dy$? $\endgroup$ – user3141592 Feb 5 '17 at 16:30
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    $\begingroup$ $x$ is a parameter so $x<y$ should be written $x<a$ here as $y$ will go from $a$ to $b$. $x<y$ (and $a<b$) implies $\int_a^b x\mathrm{d}y <\int_a^b y\mathrm{d}y$ is true. You obtain $x(b-a) < \frac{b^2-a^2}{2}$ which is tantamount to $2x<a+b$, and the last assertion is true as $x<y$ with $y\in[a,b]$. $\endgroup$ – Babyblog Feb 5 '17 at 16:37
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    $\begingroup$ But in your example $x$ is not less than $y$ for all $y \in [0,1]$ which is the interval of integration.You need to have $x<y$ for all $y\in [a,b]$... $\endgroup$ – Babyblog Feb 5 '17 at 16:52

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