MathDB

1

Part of 1991 Poland - Second Round

Problems(1)

prod a_i + prod b_i <= prod c_i + prod d_i

Source: Polish MO Recond Round 1991 p1

9/9/2024
The numbers ai a_i , bi b_i , ci c_i , di d_i satisfy the conditions 0≤ci≤ai≤bi≤di 0\leq c_i \leq a_i \leq b_i \leq d_i and ai+bi=ci+di a_i+b_i = c_i+d_i for i=1,2,…,n i=1,2 ,\ldots,n. Prove that ∏i=1nai+∏i=1nbi≤∏i=1nci+∏i=1ndi \prod_{i=1}^n a_i + \prod_{i=1}^n b_i \leq \prod_{i=1}^n c_i + \prod_{i=1}^n d_i
algebrainequalitiesProduct