Gosh, what a troll problem. The inequality is so strong that in fact, equality always holds. \[ \sum_{i = 1}^k \frac{a_i^2}{a_i B + b_i A} = \frac{1}{B} \cdot \sum_{i = 1}^k \left( a_i - \frac{a_i b_i A}{a_i B + b_i A} \right) = \frac{1}{B} \cdot A \left( 1 - \sum_{i = 1}^k \frac{a_i b_i}{a_i B + b_i A} \right) \]Similarly, \[ \sum_{i = 1}^k \frac{b_i^2}{a_i B + b_i A} = \frac{B}{A} \cdot \left( 1 - \sum_{i = 1}^k \frac{a_i b_i }{a_i B + b_i A} \right) \]Multiplying the above two identity, we conclude that \[ \left( \sum_{i = 1}^k \frac{a_i^2}{a_i B + b_i A} \right) \left( \sum_{i = 1}^k \frac{b_i^2}{a_i B + b_i A} \right) = \left( 1 - \sum_{i = 1}^k \frac{a_i b_i }{a_i B + b_i A} \right)^2 \]

$$\left( \sum_{i = 1}^{k} \frac{a_ib_i}{a_iB + b_iA} - 1 \right)^2 = \left( \sum_{i = 1}^{k} \frac{a_ib_i}{a_iB + b_iA} - 1 \right) \left( \sum_{i = 1}^{k} \frac{a_ib_i}{a_iB + b_iA} - 1 \right)$$ Since $A = \sum_{i = 1}^{k} a_i$ and $B = \sum_{i = 1}^{k} b_i$. Clearly $\sum_{i = 1}^{k} \frac{a_i}{A} = \sum_{i = 1}^{k} \frac{b_i}{B} = 1$ So, $$ = \left( \sum_{i = 1}^{k} \frac{a_ib_i}{a_iB + b_iA} - \frac{a_i}{A} \right) \left( \sum_{i = 1}^{k} \frac{a_ib_i}{a_iB + b_iA} - \frac{b_i}{B} \right) = \left( \sum_{i = 1}^{k} \frac{-(a_i)^2 B}{A (a_iB + b_iA)} \right) \left( \sum_{i = 1}^{k} \frac{-(b_i)^2 A}{B (a_iB + b_iA)} \right)$$$$ = \left( - \frac{B}{A} \left( \sum_{i = 1}^{k} \frac{a_i^2}{a_iB + b_iA} \right) \right) \left( - \frac{A}{B} \left( \sum_{i = 1}^{k} \frac{b_i^2}{a_iB + b_i} \right) \right) = \left( \sum_{i = 1}^{k} \frac{a_i^2}{a_iB + b_iA} \right) \left( \sum_{i = 1}^{k} \frac{b_i^2}{a_iB + b_iA} \right) $$

Yeah,this trolled me so hard,that I proved in the actual contest that the reverse inequality is true,as I assumed it was a mistake by the Jury.My solution can be easiely modified to prove that they are actually equal,but I don't know how many points I'll get