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$Ta có:a^{2}.\sqrt{bc}\leq a^{2}.\frac{b+c}{2}$
$b^{2}.\sqrt{ac}\leq b^{2}.\frac{a+c}{2}$
$c^{2}.\sqrt{ab}\leq c^{2}.\frac{a+b}{2}$
$Suy ra:$
$a^{2}.\sqrt{bc}+b^{2}.\sqrt{ac}+c^{2}.\sqrt{ab}\leq a^{2}.\frac{b+c}{2}+ b^{2}.\frac{a+c}{2}+c^{2}.\frac{a+b}{2}$
$<=> a^{2}.\sqrt{bc}+b^{2}.\sqrt{ac}+c^{2}.\sqrt{ab}\leq \frac{ab.(a+b)}{2}+\frac{bc.(b+c)}{2}+\frac{c.a(c+a)}{2}(*)$
$Biến đổi tương đương dễ dàng chứng minh: ab.(a+b)\leq a^{3}+b^{3} ,bc.(b+c) \leq c^{3}+b^{3},ca.(c+a) \leq a^{3}+c^{3}(**)$
$Từ(*) và (**) Suy ra : a^{2}.\sqrt{bc}+b^{2}.\sqrt{ac}+c^{2}.\sqrt{ab}\leq a^{3}+b^{3}+ c^{3}$
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