If $\prod_{i=1}^{k}a_{i}\le\prod_{i=1}^{k}b_{i}$ for $k=1,2,\ldots,n$, then show that $\prod_{k=1}^{n}(1+a_{k})\le\prod_{k=1}^{n}(1+b_{k})$. Let $a_{1}\ge a_{2}\ge\cdots\ge a_{n}\ge 0$ and $b_{1}\ge b_{2}\ge\cdots\ge b_{n}\ge 0$ be such  that
$$\prod_{i=1}^{k}a_{i}\le\prod_{i=1}^{k}b_{i}\text{ for every } k=1,2,\ldots,n\,.$$
Show that
$$\prod_{k=1}^{n}(1+a_{k})\le\prod_{k=1}^{n}(1+b_{k})\,.$$
I think can use induction to prove it. 
since $n=1$ is clear.when $n=2$ it is  must show that
$$a_{1}+a_{2}+a_{1}a_{2}\le b_{1}+b_{2}+b_{1}b_{2}$$
by condtion we have
$$a_{1}\le b_{1},~~~~~a_{1}a_{2}\le b_{1}b_{2}$$
But we can't known $a_{2}$ and $b_{2}$ which is bigger?
 A: If $a_p>0$ and $a_{p+1}=a_{p+2}=\ldots=a_n=0$, then we have $b_p>0$ and $b_{j}\geq 0$ for all $j=p+1,p+2,\ldots,n$.  Thus, if we can show that $$\prod_{i=1}^p\,\left(1+a_i\right)\leq \prod_{i=1}^p\,\left(1+b_i\right)\,,$$
then it follows immediately that
$$\prod_{i=1}^n\,\left(1+a_i\right)=\prod_{i=1}^p\,\left(1+a_i\right)\leq \prod_{i=1}^p\,\left(1+b_i\right)\leq \prod_{i=1}^n\,\left(1+b_i\right)\,.$$
Consequently, it suffices to assume that $a_1\geq a_2\geq \ldots \geq a_n>0$ and $b_1\geq b_2\geq \ldots\geq b_n>0$.
Now, consider the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x):=\ln\big(1+\exp(x)\big)$ for all $x\in\mathbb{R}$.  Note that $f$ is a strictly convex, strictly increasing function.  The sequence $\big(\ln(a_1),\ln(a_2),\ldots,\ln(a_n)\big)$ is majorized by the sequence $\big(\ln(b_1),\ln(b_2),\ldots,\ln(b_n)\big)$ in the weak sense (that is, we have $\sum\limits_{i=1}^k\,\ln(a_i)\leq \sum\limits_{i=1}^k\,\ln(b_i)$ for all $k=1,2,\ldots,n$, instead of an equality when $k=n$).  Thus, by the first remark of Karamata's Inequality, we get
$$\sum_{i=1}^n\,f\big(\ln(a_i)\big)\leq \sum_{i=1}^n\,f\big(\ln(b_i)\big)\,.$$
Ergo,
$$\prod_{i=1}^n\,\left(1+a_i\right)\leq \prod_{i=1}^n\,\left(1+b_i\right)\,.$$
The equality holds iff $a_i=b_i$ for all $i=1,2,\ldots,n$.
