parmenides51 26.08.2024 16:22 If $a$ and $b$ are positive numbers, prove the inequality $$a^2 +ab+b^2\ge 3(a+b-1).$$
tait1k27 27.08.2024 16:33 We have \begin{align*} a^2+ab+b^2-3(a+b-1) &= \left(a+\frac{b}{2}-\frac{3}{2} \right)^2+\frac{3}{4}(b-1)^2\\ &\ge 0. \end{align*}