Question

a+c/b+a+b/c+b+c/a >6.Cum se rezolva această inecuațieee???

Answer 1

[tex]\it \dfrac{a+c}{b} +\dfrac{a+b}{c} +\dfrac{b+c}{a} = \dfrac{a}{b} +\dfrac{c}{b} +\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a} = \\\;\\ =( \dfrac{a}{b}+\dfrac{b}{a})+(\dfrac{b}{c}+\dfrac{c}{b})+(\dfrac{c}{a}+\dfrac{a}{c}) \geq 2+2+ 2 = 6[/tex]


$\it \dfrac{a}{b} + \dfrac{b}{a} \geq2 \Leftrightarrow a^2+b^2\geq 2ab \Leftrightarrow a^2+b^2 - 2ab \geq 0 \Leftrightarrow (a -b)^2\geq0 \ (A)$