Question

PROBLEMA-DESIGUALDADES E INECUACIONES

Answer 1

Respuesta:

$A> 8$

Explicación paso a paso:

$Sea \: \: \: \: a,b,c \in R^{+} \: \: \: \: \: y \: \: \: \: \: a \neq b \neq c$

$(a-c)^2 > 0 \: \: \: , \: \: \: {(a - b)}^{2} > 0 \: \: \: , \: \: \: {(b - c)}^{2} > 0$

$b(a-c)^2 > 0 \: \: \: , \: \: \: c {(a - b)}^{2} > 0 \: \: \: , \: \: \: a{(b - c)}^{2} > 0$

Sumando:

$b(a-c)^2 + c {(a - b)}^{2} + a{(b - c)}^{2} > 0$

Desarrollar los binomios

$b( {a}^{2} - 2ac + {c}^{2} ) + c {( {a}^{2} - 2ab + {b}^{2} )} + a{( {b}^{2} - 2bc + {c}^{2} )} > 0$

${a}^{2}b - 2abc + {c}^{2}b+ {a}^{2}c - 2abc+ {b}^{2}c + {b}^{2} a - 2abc + {c}^{2}a > 0$

${a}^{2}b - 2abc + {c}^{2}b+ {a}^{2}c - 2abc+ {b}^{2}c + {b}^{2} a - 2abc + {c}^{2}a + 8abc> 8abc$

${a}^{2}b + {c}^{2}b+ {a}^{2}c + {b}^{2}c + {b}^{2} a + {c}^{2}a +2abc> 8abc$

${a}^{2}b + {b}^{2} a + {c}^{2}b+ {c}^{2} a + {a}^{2}c +abc+ abc+{b}^{2}c> 8abc$

$ab(a + b)+ {c}^{2}(a + b) + ac (a + b)+ bc(a + b) > 8abc$

$(a + b)(ab+ {c}^{2} + ac+ bc)> 8abc$

$(a + b)(ab+ bc + ac+{c}^{2} )> 8abc$

$(a + b)(b(a+ c) + c( a+ c))> 8abc$

$(a + b)(b + c)(a+ c)> 8abc$

$\frac{(a + b)(b + c)(a+ c)}{abc}> 8$

La afirmación correcta es:

$\frac{(a + b)(b + c)(a+ c)}{abc}> 8$

$A> 8$