Considering,


1. the set of all n dim. vectors $\{x\_i\}\_{i=1,...,n} $ such that $x\_i \geq 0 $ and $\sum\_{i=1}^{n}x\_i = K$
2. Any continuous and strictly increasing function $f^+(x)$ : $ \mathbb R^+ \to \mathbb R^+ $ (i.e. where $f^+(x+\epsilon)> f^+(x) \quad \forall \epsilon >0 \quad x \in \mathbb R^+$)


**My question:**
Is it true in general that $ \sum\_{i=1}^n x\_if(x\_i)\geq Kf(\frac{K}{n})$ for any $\{x\_i\}$ respecting 1. and $f^+$ ?


Using cauchy-schwarz ineq. I was able to prove this result for linear functions $f^+(x)=ax+b, \quad a,b>0$ , then:


$\sum\_{i=1}^n x\_if(x\_i)=Kb+a\sum\_{i=1}^n x\_i^2\geq Kb+\frac{aK^2}{n}$ since $n\sum\_{i=1}^n x\_i^2\geq (\sum\_{i=1}^n x\_i)^2=K^2$ .


(where $n\sum\_{i=1}^n x\_i^2=K^2$ if $x\_i=K/n$)


Intuitively I was thinking that this should lead to a more general proof of the statement since for any interval $I=[x\_{min},x\_{max}]$, we can always find $a,b>0$ such that $ax+b\leq f^+(x) \quad \forall x \in I$. But however I didn't find yet the way to conclude using this path.


I am missing something? Is there another way to proceed?


Thank's for your help!



Lower bound for $ \sum_{i=1}^n x_i f(x_i)$ when $\sum_{i=1}^{n}x_i = K$
Actually,the conclusion is negative.


By **chebyshev's theorem**, we have 
$$
\sum\_{i=1}^{n}{x\_{i}\cdot f(x\_{i})}\geq \frac{\sum\_{i}^{n}{f(x\_{i} ) } }{n} \times \sum\_{1}^{n}{x\_{i} } 
$$ 
$$
\forall x\_{1}<x\_{2}<...<x\_{n}\in \left\{ x\_{i} \right\} ,
f^{+}(x+\epsilon ) >f^{+}(x+ ) ,\forall \epsilon >0,x\in R^+
$$ 


By **Jensen**, if $f(x)$ is convex, we have 
$$
\frac{\sum\_{i}^{n}{f(x\_{i} ) } }{n}\times \sum\_{1}^{n}{x\_{i} } \geq f(\frac{\sum\_{1}^{n}{x\_{i} } }{n} )\times \sum\_{1}^{n}{x\_i} 
$$
$$
\forall x\_{1}<x\_{2}<...<x\_{n}\in \left\{ x\_{i} \right\} ,
f^{+}(x+\epsilon ) >f^{+}(x+ ) ,\forall \epsilon >0,x\in R^+
$$  

So,if $f(x)$ is convex,we have
$$
\sum\_{i=1}^{n}{x\_{i}\cdot f(x\_{i})}\ \geq f(\frac{\sum\_{1}^{n}{x\_{i} } }{n} )\times \sum\_{1}^{n}{x\_i} 
$$
(only *Jensen* is not enough to prove this)


