Maths Olympiad Prep

Library / Ukrainian National Mathematical Olympiad /34 of 201

Ukrainian National Mathematical Olympiad

Algebra Difficulty 5.2 AIME, harder Prove it Ukraine

What is the smallest possible value the expression ab+a+bab + a + b can take, if real numbers a,ba, b satisfy the condition a2+b2=25a^2 + b^2 = 25.

Solution

(a+b+1)2≥0(a+b+1)^2 \ge 0
a2+b2+1+2ab+2a+2b≥0⇒ab+a+b≥−12(a2+b2+1)=−13.a^2 + b^2 + 1 + 2ab + 2a + 2b \ge 0 \Rightarrow ab + a + b \ge -\frac{1}{2}(a^2 + b^2 + 1) = -13.
By choosing a=−4a = -4 and b=3b = 3, we obtain: ab+a+b=−12−4+3=−13ab + a + b = -12 - 4 + 3 = -13, i.e. the smallest possible value is reached.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.