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Sa se demonstreze ca \( 1+\frac{3}{a+b+c}\ge \frac{6}{ab+bc+ca} \) , cu \( a,b,c>0 \) si \( abc=1 \) .

Notind \( x=\frac{1}{a} \) , \( y=\frac{1}{b} \) , \( z=\frac{1}{c} \) atunci \( xyz=1 \) si inegalitatea devine

\( 1+\frac{3}{xy+yz+zx}\ge\frac{6}{x+y+z} \) care este evidenta deoarece

\( LHS\ge 2\sqrt{\frac{3}{xy+yz+zx}}\ge2\cdot\frac{3}{x+y+z} \)

\( (ab + bc + ca)^2\ge 3abc(a + b + c) = 3(a + b + c) \)

Deci \( 1 + \frac {3}{a + b + c} \geq 1 + \frac {9}{(ab + ac + bc)^2}\geq \frac {6}{ab + ac + bc} \)