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Fie \( a,b,c,d \) patru numere reale nenegative . Sa se arate ca inegalitatea urmatoare este adevarata :

\(
\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+a}+\frac{d-a}{d+2a+b}\ge 0 \)

Scriem inegalitatea sub forma \( \sum\left\(\frac{a-b}{a+2b+c}+\frac{1}{2}\right\)\ge 2 \)

\( \Longleftrightarrow \sum\frac{3a+c}{a+2b+c}\ge 4. \)

Din inegalitatea Cauchy-Schwarz avem

\( \sum\frac{3a+c}{a+2b+c}\ge \frac{\[\sum(3a+c)\]^2}{\sum(3a+c)(a+2b+c)}=\frac{16(a+b+c+d)^2}{4(a+b+c+d)^2}=4 \)