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Fie \( a,b,c\in\left(0,\infty\right) \) astfel incat \( b^{2}<ac \) si \( c^{2}<ab \). Sa se demonstreze ca \( \frac{bc}{ac-b^{2}}+\frac{ac}{ab-c^{2}}\ge\frac{4ab}{a^{2}-bc} \).

Emil C. Popa, concursul "Gh. Lazar", 2009

Indicatie: \( \frac{a^2}{x}+\frac{b^2}{y}\ge \frac{(a+b)^2}{x+y} \) , \( x,y\ge 0 \)

Inegalitatea se mai scrie sub forma:
\( \frac{1}{\frac{ac}{bc}-\frac{b^2}{bc}}+\frac{1}{\frac{ab}{ac}-\frac{c^2}{ac}}\ge \frac{4}{\frac{a^2}{ab}-\frac{bc}{ab}} \) sau \( \frac{1}{\frac{a}{b}-\frac{b}{c}}+\frac{1}{\frac{b}{c}-\frac{c}{a}}\ge \frac{4}{\frac{a}{b}-\frac{c}{a}} \).
Folsind inegalitatea lui Cauchy scrisa sub forma \( \frac{a^2}{x}+\frac{b^2}{y}\ge \frac{(a+b)^2}{x+y} \) obtinem: \( \frac{1}{\frac{a}{b}-\frac{b}{c}}+\frac{1}{\frac{b}{c}-\frac{c}{a}}\ge \frac{(1+1)^2}{\frac{a}{b}-\frac{b}{c}+\frac{b}{c}-\frac{c}{a}}=\frac{4}{\frac{a}{b}-\frac{c}{a}} \), ceea ce incheie demonstratia.