Algebra · Inequalities · Arithmetic and geometric mean · Algebraic identities · Square roots
Let \(a\) and \(b\) be real numbers with \(0 < b \leqslant a\). Prove that
\[ \frac{1}{8} \cdot \frac{(a-b)^{2}}{a} \;\leqslant\; \frac{a+b}{2} - \sqrt{ab} \;\leqslant\; \frac{1}{8} \cdot \frac{(a-b)^{2}}{b} . \]
Determine under which conditions one of these inequalities becomes an equality.
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Serbian Regional Competition 2007, high school grade I, category A, problem 4. Organized by the Mathematical Society of Serbia (DMS). Source