# Prove $e^{a_1}+...+e^{a_n}\leqslant e^{b_1}+...+e^{b_n}$, where $a_1+...+a_n=b_1+...+b_n=0$, $|a_j|\leqslant |b_j|$

Let $\{a_1,\ldots,a_n\}$ and $\{b_1,\ldots,b_n\}$ $-$ sets of real numbers ($n\in \mathbb{N}$), such that $$a_1+\cdots+a_n = b_1+\cdots+b_n=0, \qquad |a_j|\leqslant |b_j|, \; j=1,\ldots,n,$$
$a_j$ and $b_j$ $-$ of same sign ($a_j\cdot b_j\geqslant 0$).

Prove inequality

$$e^{a_1}+\cdots+e^{a_n} \leqslant e^{b_1}+ \cdots+e^{b_n}.$$

• Of course, in nontrivial case some $a_j$ ($b_j$) are positive/negative. Jun 25, 2013 at 9:44
• Did you try using induction?
– SBF
Jun 25, 2013 at 9:48
• @Ilya, I had difficulties with [induction] + [$a_1+...+a_n=b_1+...+b_n$]. Each step must correct previous statement. I think must be elegant way of proof. Jun 25, 2013 at 9:55

Here is a proof by induction on the number of $j$ such that $a_j\neq b_j$. For the base case, if $a_j=b_j$ for all $j$ we are done.

Otherwise, given $(a_1,\dots,a_n)\neq (b_1,\dots,b_n)$ satisfying your conditions, we want to show $e^{a_1}+\dots+e^{a_{n}}\leq e^{b_1}+\dots+e^{b_{n}}$. Since $\sum_{i=1}^n (a_i-b_i)=0$, there are indices $i,j$ such that $|a_i|<|b_i|$ and $|a_j|<|b_j|$ with $a_i>0$ and $a_j<0$. Let $\delta=\min(b_i-a_i,a_j-b_j)$ and consider the modified sequence defined by $a'_i=a_i+\delta$ and $a'_j=a_j-\delta$ and $a'_k=a_k$ for all $k\notin\{i,j\}$. Note that $$e^{a_i}+e^{a_j}\leq e^{a_i+\delta}+e^{a_j-\delta} = e^{a'_i}+e^{a'_j}.$$ because the function $e^{x+\delta}-e^x$ is nondecreasing in $x$ and $a_j<0<a_i$. Therefore $e^{a_1}+\dots+e^{a_{n}}\leq e^{a'_1}+\dots+e^{a'_{n}}$. But by the induction hypothesis we have $e^{a'_1}+\dots+e^{a'_{n}}\leq e^{b_1}+\dots+e^{b_{n}}$.

First, exclude all the terms such that $a_i=b_i$, since these obviously do not affect the inequality. We show that the sequences can be made so that $(b_i)$ majorises $(a_i)$, and then we apply Karamata's inequality.

Let $a_1,b_1,\cdots a_p,b_p\geq 0$ and the remaining terms negative, and the same for $b_i.$ Since $|a_i|<|b_i|:$

$$a_1< b_1$$

$$a_1+a_2< b_1+b_2$$

$$\cdots$$

$$a_1+\cdots +a_p< b_1+\cdots +b_p$$

There are two possibilities here. Either $n=p+1,$ in which case adding $a_{p+1},b_{p+1}$ yields equality and we are done. Or $n>p+1$, in which case we must still have:

$$a_1+\cdots+a_{p+1}<b_1+\cdots +b_{p+1}$$

This follows from a simple contradiction: suppose $a_1+\cdots+a_{p+1}\geq b_1+\cdots +b_{p+1}$ then $a_1+\cdots +a_{n}>b_1+\cdots+b_n$ since $a_j>b_j$ for $j>p$. This is false, hence inductively:

$$a_1+\cdots +a_i\leq b_1+\cdots +b_i,\;\;1\leq i\leq n$$

Thus $(b_i)$ majorises $(a_i)$, and for any convex $f:$

$$f(a_1)+f(a_2)+\cdots +f(a_n)\leq f(b_1)+f(b_2)+\cdots +f(b_n)$$

This problem is simply the case $f=\exp$.

• Thank you for link to Karamata's inequality! very useful. Jun 25, 2013 at 13:37
• It was the most elegant proof (as for me). Jun 25, 2013 at 21:22

Consider 2 sets of indices: $J_{+}: j, \mbox{for which} a_j\geqslant 0$, $\quad$ $J_{-}: j, \mbox{for which} a_j< 0$. $\;\;$ Then $$-\sum_{j\in J_-} a_j = \sum_{j\in J_+} a_j, \qquad -\sum_{j\in J_-} b_j = \sum_{j\in J_+} b_j. \tag{1}$$

If $j\in J_+$, then $0\leqslant a_j \leqslant b_j$ (based on Mean value theorem for $f(x)=e^x$ ), there exists $c_j \in [a_j,b_j]$: $$e^{b_j} - e^{a_j} = (b_j-a_j) e^{c_j} \geqslant b_j-a_j.\tag{2}$$

If $j\in J_-$, then $b_j \leqslant a_j\leqslant 0$, then there exists $c_j \in [b_j,a_j]$: $$e^{a_j} - e^{b_j} = (a_j-b_j) e^{c_j} \leqslant a_j-b_j;$$ $$e^{b_j} - e^{a_j} = (a_j-b_j) e^{c_j} \geqslant b_j-a_j. \tag{3}$$

$(2),(3),(1)\implies$

$$\sum_{j=1}^n (e^{b_j}-e^{a_j}) = \sum_{j\in J_+}(e^{b_j}-e^{a_j}) + \sum_{j\in J_-} (e^{b_j}-e^{a_j})$$ $$\geqslant \sum_{j\in J_+}(b_j-a_j) + \sum_{j\in J_-}(b_j-a_j) = 0.$$

It was my proof.
Here positive and negative cases are considered apart.
I'm looking for more elegant proof.