MathDB
a^2+a+b^2>= a^4+a^3+b^4 - 2008 Cuba MO 2.7

Source:

August 27, 2024
algebrainequalities

Problem Statement

For non negative reals a,ba,b we know that a2+a+b2a4+a3+b4a^2+a+b^2\ge a^4+a^3+b^4. Prove that 1a4a2b21b\frac{1-a^4}{a^2}\ge \frac{b^2-1}{b}