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In fact it's my point of view but I think that this version of Niculescu's inequality is a mixture between Karamata and Jensen's inequality .

So we have :

Let $f(x)$ be a convex strictly decreasing function on an interval $I$, $x_i$ and $y_i$ be positives reals numbers belonging to $I$ , and $p_i>0$ be the coefficients such that $\sum_{i=1}^{n}p_i=1$ then we have : $$\sum_{i=1}^{n}p_if(y_i)-f(\sum_{i=1}^{n} p_i y_i)\geq \sum_{i=1}^{n}p_if(x_i)-f(\sum_{i=1}^{n} p_i x_i) $$

With the Condition $$\sum_{i=1}^{k}p_i x_i\geq \sum_{i=1}^{k}p_i y_i$$ With : $1\leq k \leq n$

My First question : Is this theorem true ? And have some literature on this ?

If it's true I have a proof for the following inequality :

$$j(\frac{x_1^{q+1}}{k x_1^{q}+ p x_2^{q}}+\cdots+\frac{x_n^{q+1}}{k x_n^{q}+ p x_1^{q}})\geq \sum_{i=1}^{n} x_i$$ With $j=k+p$ positive real numbers and $q$ a natural positive number .

My try :

Let the following substitution coming :

$$a_i=\frac{x_{i+1}}{x_i}$$

After simplify (dividing) the inequality by a $x_i$ ( you have $n$ choice ) the form of the inequality becomes :

$$j(\frac{\alpha_1}{k + p a_1^{q}}+\cdots+\frac{\alpha_n}{k + p a_n^{q}})\geq \sum_{i=1}^{n} \alpha_i$$

With $\alpha_1=1$

$\alpha_2=a_1$ and

$\alpha_k=\prod_{i=1}^{k-1} a_i$

Now we divide by $\sum_{i=1}^{n} \alpha_n$ we get :

$$j(\frac{p_1}{k + p a_1^{q}}+\cdots+\frac{p_n}{k + p a_n^{q}})\geq 1$$ With $$p_i=\frac{\alpha_i}{\sum_{i=1}^{n} \alpha_i}$$ Now we apply the theorem of Niculescu to the function :

$-F(x)$ such that :

$$F'(x)=j(\frac{1}{k + p x^{q}})$$

We know that $F(x)$ is increasing and concave .

With :

$x_i=a_i+\epsilon$

$y_i=a_i$

And

$p_i=\frac{\alpha_i}{\sum_{i=1}^{n}\alpha_i}$

Now remains to divide by epsilon the inequality and make the difference between each summand with the same index and take the limit for $\epsilon$ to zero and apply the Fondamental theorem of Calculus .I let the end to the reader .

My last question : Is my proof right ?

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