Toán Lớp 9: Cho a , b , c > 0. Chứng minh :
a^3 + b^3 + abc >= ab(a+b+c)
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Toán Lớp 9: Cho a , b , c > 0. Chứng minh :
a^3 + b^3 + abc >= ab(a+b+c)
Comments ( 2 )
$⇔a^3+b^3+abc≥a^2b+ab^2+abc$
$⇔a^3+b^3-a^2b-ab^2≥0$
$⇔a^2(a-b)-b^2(a-b)≥0$
$⇔(a-b)(a^2-b^2)≥0$
$⇔(a-b)^2(a+b)≥0$ (Luôn đúng với $∀a,b>0)$
$⇒a^3+b^3+abc≥ab(a+b+c)∀a,b,c>0(đpcm)$