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Fie \( a,b,c>0 \) . Sa se demonstreze ca :

\( \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c+\frac{4(a-b)^2}{a+b+c} \)

Avem ca :

\( \frac {a^{2}}{b} + \frac {b^{2}}{c} + \frac {c^{2}}{a} - (a + b + c) = \frac {(a - b)^{2}}{b} + \frac {(b - c)^{2}}{c} + \frac {(c - a)^{2}}{a} \)

Aplicam inegalitatea Cauchy-Schwarz :
\( \frac {(b - c)^{2}}{c} + \frac {(c - a)^{2}}{a}\ge \frac {(b - c + c - a)^{2}}{c + a} = \frac {(a - b)^{2}}{c + a} \)

Si :

\( \frac {1}{b} + \frac {1}{c + a}\ge \frac {4}{a + b + c} \)

Deci :

\( \frac {(a - b)^{2}}{b} + \frac {(b - c)^{2}}{c} + \frac {(c - a)^{2}}{a}\ge (a - b)^{2}(\frac {1}{b} + \frac {1}{c + a}) \ge \frac {4(a - b)^{2}}{a + b + c} \)

Egaliatea are loc daca si numai daca \( a = b = c \).