4
$\begingroup$

Related to this I have solved the general case which is the following:

If $a_1,\cdots,a_n >0$, and $a_1+\cdots + a_n=n$, prove that $$\sum_{cyc}a_i^{a_i a_{i+1}} \geq (\sum_{i=1}^{n}a_ia_{i+1})\ln\left(\frac{\sum_{i=1}^{n}a_ia_{i+1}}{n}\right)+n> n(1-\frac{1}{e})$$

Where $a_{n+1}=a_1$


My proof:

We begin with a substitution we put:

$a_i=e^{b_i}$

After this we take the theorem 1.5 of this paper and put:

$\phi(x)=e^{-x}=\frac{1}{f(x)}$

$d=0$

$p_i=\frac{1}{\phi{(x_i)}}$

$x_i=b_ie^{b_i+b_{i+1}}$

We get:

$$\frac{n}{\sum_{i=1}^{n}f(x_i)}\leq 1+\frac{\sum_{i=1}^{n} -x_i}{\sum_{i=1}^{n}f(x_i)}$$

Or:

$$n+\sum_{i=1}^{n}x_i\leq \sum_{i=1}^{n}f(x_i)$$

Which is equivalent to the inequality:

$$n+\sum_{cyc}\ln(a_i^{a_i a_{i+1}})\leq \sum_{cyc}a_i^{a_i a_{i+1}}$$

Now we use Jensen inequality to give:

$$\sum_{cyc}\ln(a_i^{a_i a_{i+1}})\geq (\sum_{i=1}^{n}a_ia_{i+1})\ln\left(\frac{\sum_{i=1}^{n}a_ia_{i+1}}{n}\right)$$

So we get:

$$\sum_{cyc}a_i^{a_i a_{i+1}} \geq (\sum_{i=1}^{n}a_ia_{i+1})\ln\left(\frac{\sum_{i=1}^{n}a_ia_{i+1}}{n}\right)+n>n(1-\frac{1}{e})$$


My question is:

Can you check my proof and tell me if there exist a better constant for the original inequality?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy