We let $p=rank(AB)$, since $p \leq min(rank(A),rank(B))$, it suffices to prove that $$\sum_{i=1}^p\sigma_i(AB) \leq \sum_{i=1}^p\sigma_i(A)\sigma_i(B).$$
Consider the SVD decomposition of AB,
$$AB=\left[\begin{array}{cc}U & \widetilde{U} \end{array}\right]\left[\begin{array}{cc} \Sigma & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{c} V^T \\ \widetilde{V}^T \end{array}\right]=U\Sigma V^T$$
where $\Sigma=diag(\sigma_1(AB),\ldots,\sigma_p(AB)).$
For any matrix $P$, we let $P_k$ denotes the submatrix consisting of the first $k$ columns of $P$. We fix a $k \in \left\{ 1,\ldots,p\right\},$
$$U_k^T(AB)V_k=diag(\sigma_1(AB),\ldots,\sigma_k(AB)).$$
Next, we consider the SVD of $U_k^TA,$
$$U_k^TA=R \left[\begin{array}{cc}S & 0 \end{array}\right]\left[\begin{array}{c} W^T\\ \widetilde{W}^T\end{array} \right]=RSW^T.$$
We have
$$U_k^TAA^TU_k=RS^2R^T$$
and hence we can see that
$$\det(RS^2R^T)\leq\prod_{i=1}^k \sigma_i(A)^2$$
and as a result,
$$\det(RSR^T) \leq \prod_{i=1}^k \sigma_i(A).$$
\begin{align*}
\prod_{i=1}^k \sigma_i(AB) &=\det(U_k^TABV_k)\\
&= \det(RSW^TBV_k) \\
&= \det(RSR^T)det(RW^TBV_k)\\
&\leq \prod_{i=1}^k \sigma_i(A)\sigma_i(B).
\end{align*}
Taking logarithm on both sides, we have $\forall k \in \left\{1, \ldots, p\right\},$
$$\sum_{i=1}^k \log(\sigma_i(AB)) \leq \sum_{i=1}^k \log(\sigma_i(A)\sigma_i(B))$$
I will use a trick called majorization to remove the logarithms. I am going to construct 2 vectors $a,b \in \mathbb{R}^{p+1}$ such that $a \succ b.$ The vectors satisfy the following conditions:
\begin{align*}
\text{1. }& a_1 \geq \ldots \geq a_{p+1},\\
\text{2. }& b_1 \geq \ldots \geq b_{p+1}, \\
\text{3. }& \sum_{i=1}^k b_i \leq \sum_{i=1}^k a_i, \forall k \in \left\{1,\ldots,p\right\}, \\
\text{4. }& \sum_{i=1}^{p+1} b_i = \sum_{i=1}^{p+1} a_i.
\end{align*}
Under the above conditions, $\sum_{i=1}^{p+1} \exp(b_i) \leq \sum_{i=1}^{p+1} \exp(a_i)$ since the exponential function is convex.
We let
$$b=\left(\log(\sigma_1(AB),\ldots,\log(\sigma_p(AB)),\min(a_p,b_p)\right),$$
$$a=\left(\log(\sigma_1(A)\sigma_1(B),\ldots,\log(\sigma_p(A)\sigma_p(B)),\sum_{i=1}^{p+1}b_i-\sum_{i=1}^p a_i\right).$$
To verify the $\textbf{Condition 1}$,
I need to show that $a_p \geq \sum_{i=1}^{p+1}b_i-\sum_{i=1}^p a_i$ which is equivalent to
$$a_p-\sum_{i=1}^{p+1}b_i+\sum_{i=1}^p a_i=(a_p-b_{p+1})+\left(\sum_{i=1}^pa_i-\sum_{i=1}^p b_i\right)\geq 0.$$
$b_{p+1} \leq a_p$ due to the definition of $b_{p+1}$ and we have proven earlier that $\left(\sum_{i=1}^pa_i-\sum_{i=1}^p b_i\right) \geq 0$.
To verify the $\textbf{Condition 2}$, observe that $b_p \geq b_{p+1}$ due to definition of $b_{p+1}$ again.
$\textbf{Condition 3}$ was proven earlier.
To check $\textbf{Condition 4}$,
$$\sum_{i=1}^{p+1}a_i=\sum_{i=1}^{p}a_i+a_{p+1}=\sum_{i=1}^{p+1}b_i.$$
As a result, we have
$$\sum_{i=1}^{p+1} \exp(b_i) \leq \sum_{i=1}^{p+1} \exp(a_i)$$
which is equivalent to
$$\sum_{i=1}^{p} \exp(b_i) \leq \sum_{i=1}^{p} \exp(a_i)+\exp(a_{p+1})-\exp(b_{p+1}).$$
Thus, we have
\begin{align*}
\sum_{i=1}^{p} \sigma_i(AB) &\leq \sum_{i=1}^{p} \sigma_i(A)\sigma_i(B)+\exp(a_{p+1})-\exp(b_{p+1})\\
&=\sum_{i=1}^{p} \sigma_i(A)\sigma_i(B)+\exp\left(\sum_{i=1}^p(b_i-a_i)+b_{p+1}\right)-\exp(b_{p+1})\\
&=\sum_{i=1}^{p} \sigma_i(A)\sigma_i(B)+\exp(b_{p+1})\left(\exp \left(\sum_{i=1}^p(b_i-a_i)\right)-1)\right)\\
&\leq \sum_{i=1}^{p} \sigma_i(A)\sigma_i(B)+\exp(b_{p+1})\left(\exp \left(0\right)-1)\right)\\
&=\sum_{i=1}^{p} \sigma_i(A)\sigma_i(B)
\end{align*}
Hence we are done.
A bonus that I learn from answering the question is if the following conditions hold:
\begin{align*}
\text{A. }& a_1 \geq \ldots \geq a_{p},\\
\text{B. }& b_1 \geq \ldots \geq b_{p}, \\
\text{C. }& \sum_{i=1}^k b_i \leq \sum_{i=1}^k a_i, \forall k \in \left\{1,\ldots,p\right\}, \\
\end{align*}
Then we have $$\sum_{i=1}^k \exp(b_i) \leq \sum_{i=1}^k \exp(a_i), \forall k \in \left\{1,\ldots,p\right\}.$$
It should be $a^5 - a^2 + 3$ etc. rather than $a^5 - a^2 + 2$ etc.
Problem: Let $a, b, c $ be positive reals. Prove that
$$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3.$$
Proof: It is easy to prove that $x^5 - x^2 + 3 \ge x^3 + 2$ for all $x \ge 0$.
It suffices to prove that
$$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3.$$
Using Holder's inequality, we have
\begin{align*}
(a^3+2)(b^3+2)(c^3+2)
&= (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3)\\
&\ge (\sqrt[3]{a^3 \cdot 1 \cdot 1} + \sqrt[3]{1 \cdot b^3 \cdot 1} + \sqrt[3]{1\cdot 1 \cdot c^3})^3\\
& = (a + b + c)^3.
\end{align*}
For Holder's inequality, see:
https://artofproblemsolving.com/wiki/index.php/H%C3%B6lder%27s_Inequality
Best Answer
Assume $a_i=b_i={1\over n}.$ Then $$\left(\frac{a_i}{\sum_{i =1}^n a_i}\right)^\alpha \left(\frac{b_i}{\sum_{i=1}^n b_i}\right)^\beta={1\over n}.$$ Let $c_i={1\over ni}.$ Then $${c_1\over \sum_{i=1}^n c_i} < {1\over n}< {c_n\over \sum_{i=1}^n c_i}$$ Therefore the desired inequality is not satisfied for $n\ge 2.$