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Jensen's inequality states that for a convex function $f:I \to \mathbb R$ and $x_1,...,x_n\in I$, we have $$\frac{1}{n}\sum_{i=1}^n f(x_i)\ge f\left(\frac{1}{n}\sum_{i=1}^n x_i\right)$$ But in order to use this inequality, you have to choose the right function. Take for example this problem

$x_1,...,x_n$ are positive reals such that $$\sum_{i=1}^n\frac{1}{1+x_i}=1$$ Show that $$\prod_{i=1}^nx_i\ge (n-1)^n$$

Here the function that we're looking for turns out to be $f(x)=\frac{1}{1+e^x}$

In the proof of the AM-GM inequality, we've used $f(x)=\ln x$.

So is there a way to know what's the function that will help me solve my inequality?

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  • $\begingroup$ What is your inequality? $\endgroup$
    – Crostul
    Mar 1, 2022 at 20:23
  • $\begingroup$ @Crostul I'm talking in general. If you have an inequality how can you determine $f$ $\endgroup$
    – PNT
    Mar 1, 2022 at 20:32
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    $\begingroup$ I dont' think you'll find satisfactory answers if you are not specific. You can't find a "general function" for a "general inequality". $\endgroup$
    – Crostul
    Mar 1, 2022 at 20:34
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    $\begingroup$ @Crostul I'm not asking for a general function. I just want some ideas that will help me find $f$ providing some examples. Something I've just discovered right now is that if you have a sum that is $\ge$ or $\le$ than a product, you should probably include $e^x$ or $\ln x$ in your function so that you can use properties of logarithms. $\endgroup$
    – PNT
    Mar 1, 2022 at 20:40
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    $\begingroup$ It comes from experience / expanding your bag of tricks. Look at solutions in the jensen-inequality tag to see if you can compile some good examples. There are some good things to try, but no "hard and fast rule". Sometimes you have to play around with what function works, and tweak it slightly. $\quad$ EG In your example, think about why $ 1/(1+x)$ doesn't work as well as $ 1/ (1 + e^x)$, even though both are convex. What is the conclusion we get when we use $1 / (1+x)$? $\endgroup$
    – Calvin Lin
    Mar 8, 2022 at 14:53

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I think this is highly subjective, but I will try to answer based on my experience of solving some inequalities. Note that I am no expert by any means.

A lot of the times, using the correct function is kind of obvious. What is tricky, is figuring out the weights.

For example, as you say, the proof of AM-GM follows immediately from the convexity of $f(t) = \log(t)$. One way to see why this function works, is that the log function maps a multiplicative identity to an additive one, which is exactly what happens in the AM-GM inequality.

Another "trivial" inequality is $$ \left(\frac{x_1+\cdots+x_n}{n} \right)^p\leq \frac{x_1^p+\cdots+x_n^p}{n}, $$ which again follows immediately from Jensen when applied to the function $t^p$, $p > 1$.

In other cases, the right function to use becomes obvious only after making some arrangements of the terms. This is especially the case if one can homogenize or normalize the inequality in some way. Take for example Minkowski's inequality: $$ \left(\sum_{k=1}^n \lvert x_k + y_k\rvert^p\right)^{1/p}\leq \left(\sum_{k=1}^n \lvert x_k\rvert^p\right)^{1/p} + \left(\sum_{k=1}^n \lvert y_k\rvert^p\right)^{1/p}. $$ It can in fact be derived if you choose $w_k=|y_k|^p/\sum_{j=1}^n|y_k|^p$, and $t_k = |x_k/y_k|^p$ in Jensen's inequality for the concave function $f(t) = (1+t^{1/p})^p$. One thing that gives this function away after rearranging is dividing both sides by $\left(\sum_{k=1}^n \lvert x_k\rvert^p\right)^{1/p}$.

Holder's inequality is another example of this, as pointed out by Gauge_name.

Another interesting way of using Jensen is in this answer: https://math.stackexchange.com/a/1033415/443030. Notice how homogenization works here.

Finally, one tricky inequality proven using Jensen is $$ 2\leq \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\frac{a_3}{a_4+a_1}+\frac{a_4}{a_1+a_2}. $$ Divide both sides by $S=a_1+a_2+a_3+a_4$ and use the convexity of $f(t)=1/t$ to get $$ \frac{1}{S}\left(\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\frac{a_3}{a_4+a_1}+\frac{a_4}{a_1+a_2}\right)\geq \frac{S}{a_1(a_2+a_3)+a_2(a_3+a_4)+a_3(a_4+a_1) + a_4(a_1+a_2)}. $$ All that remains to prove is $$ \frac{S^2}{a_1(a_2+a_3)+a_2(a_3+a_4)+a_3(a_4+a_1) + a_4(a_1+a_2)} \geq 2, $$ which is easy.

In conclusion, as Calvin Lin pointed out in the comments, figuring out the correct way to use Jensen requires one to play around with the inequality until the convexity emerges. Certainly solving lots of inequalities goes a long way. However, two of the best tricks I found to work quite well is homogenization and normalization. These help a lot with figuring out the weights and values to be used in Jensen, and to a large extent helps to find a "tight" function to use.

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I don't know if you consider this to be an answer but in proving Holder inequality for example you may use Jensen's inequality and the function you choose there is $x\to x^q$ ($q>1$, $x>0$): the reason you choose this function, roughly speaking (I believe no rigorous argument can be provided in general), is that you want to estimate a product of two functions with the product of their $L^p$ and $L^q$ norms (which are the integral of power functions with exponents $p$ and $q$: this may be one reason why you choose $x^q$).

Let $(X,\mu)$ be a measure space and $f,g:X\longrightarrow \mathbb{R}$ be measurable functions, then: \begin{equation} ||fg||_{1}\leq ||f||_p ||g||_q. \end{equation}

If the r.h.s. is infinite there is nothing to prove, so assume that it is finite. Define $\nu = \frac{|f|^{p}}{||f||_p^p}\mu$: clearly $\nu$ is a probability measure. Now define $\Psi: \mathbb{R}_+\longrightarrow \mathbb{R}_+$ to be the function $\Psi(x)= x^q$. By Jensen's inequality you have

$$\Psi\biggl(\int_X g d\nu\biggr)\leq \int_X\Psi(g)d\nu$$

and in our case this means $$||fg||_1^q\leq ||f||_p^q ||g||_q^q,$$ which is our inequality.

For what concerns the inequality you proposed and in general I believe there is no rigorous argument (as I have already written), I guess it depends on the functions involved in the inequality. Even if this is not a complete answer I hope it can help you at least.

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