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[barcavodich]

$ a,b,c>0:\; abc=1 $ CMR
\[ a^3+b^3+c^3+6\ge (a+b+c)^2 \]

[Secrets In Inequalities VP]

$ a,b,c>0:\; abc=1 $ CMR
\[ a^3+b^3+c^3+6\ge (a+b+c)^2 \]

Lâu lâu ghé box BĐT chém tí !
Theo BĐT $Schur$ ta có :
$$a^{3}+b^{3}+c^{3}+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)$$
$$\Rightarrow 3(a^{3}+b^{3}+c^{3})+9abc\geq 3(ab(a+b)+bc(b+c)+ca(c+a))$$
$$\Rightarrow 4(a^{3}+b^{3}+c^{3})+15abc\geq a^{3}+b^{3}+c^{3}+3(ab(a+b)+bc(b+c)+ca(c+a))+6abc$$
$$\Rightarrow 4(a^{3}+b^{3}+c^{3})+15abc\geq (a+b+c)^3$$
$$\Rightarrow 4(a^{3}+b^{3}+c^{3})+15\geq (a+b+c)^3$$
$$\Rightarrow 4(a^{3}+b^{3}+c^{3})+24\geq (a+b+c)^3+9$$
Theo $AM-GM$ thì :
$$\frac{(a+b+c)^3}{3}+\frac{(a+b+c)^3}{3}+ \frac{(a+b+c)^3}{3}+9\geq 4\sqrt[4]{\frac{(a+b+c)^9}{3}}$$
$$ 4\sqrt[4]{\frac{(a+b+c)^9}{3}}\geq 4\sqrt[4]{\frac{(a+b+c)^8.3\sqrt[3]{abc}}{3}}= 4(a+b+c)^2$$
Do đó : $4(a^{3}+b^{3}+c^{3})+24\geq 4(a+b+c)^2$
Chia $2$ vế cho $4$ ta được ngay $Q.E.D$