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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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    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? – 2017-02-05
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    Yes, sorry. Corrected – 2017-02-05

1 Answers 1

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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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    Then, is it as easy as multiplying both sides by $dy$? And aren't there any limitations on $a$ and $b$? – 2017-02-05
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    Yes, I supposed $a – 2017-02-05
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    My real question: and why can't yo apply that to $x – 2017-02-05
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    $x$ is a parameter so $x$y$ will go from $a$ to $b$. $x$y\in[a,b]$. – 2017-02-05
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    In that example you give, let $a=0$ and $b=1$. Your last expression says that $2x<1$, hence $x<1/2$; but for $x=2/3$ and $y=3/4$, $x$y \in [a,b]$ but $x>1\2$ – 2017-02-05
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    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\in [a,b]$... – 2017-02-05
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    Oooh gotcha! That makes sense. Thank you – 2017-02-05