Bilun

Cám ơn bạn đã nhắc để mình sửa lại

$4b)a^3+2\geq 3a,b^3+2\geq 3b,c^3+2\geq 3c\Rightarrow a^3+b^3+c^3\geq a+b+c+2(a+b+c)-6;a+b+c\geq 3\sqrt[3]{abc}=3\Rightarrow \Leftrightarrow a^3+b^3+c^3\geq a+b+c$

$4c)2a^3+2b^3+2c^3+3=(2a^3+1)+(2b^3+1)+(2c^3+1)\geq 3a^2+3b^2+3c^2;a^2+b^2+c^2\geqslant 3\sqrt[3]{a^2b^2c^2}=3\Rightarrow 2a^3+2b^3+2c^3\geq 2(a^2+b^2+c^2)\Rightarrow a^3+b^3+c^3\geq a^2+b^2+c^2$

2c)$4(b+c)(c+a)\leq (a+b+2c)^2=(1+c)^2;4(a+b)(b+c)\leq (1+b)^2;4(c+a)(a+b)\leq (1+a)^2\Rightarrow 64((a+b)(b+c)(c+a))^2\leq ((1+a)(1+b)(1+c))^2\Rightarrow(1+a)(1+b)(1+c)\geq 8(a+b)(b+c)(c+a)=8(1-a)(1-b)(1-c)$

$\sum \sqrt[3]{a+b}=\sqrt[3]{\frac{9}{4}}\sum \sqrt[3]{\frac{2}{3}.\frac{2}{3}(a+b)}\leq \sqrt[3]{\frac{9}{4}}(\frac{a+b+4/3+b+c+4/3+c+a+4/3}{3})=\sqrt[3]{18}$

Bài 1.1:$(x+1)^3=4-3\sqrt[3]{2+2\sqrt{3}}.\sqrt[3]{2-2\sqrt{3}}(\sqrt[3]{2+2\sqrt{3}}+\sqrt[3]{2-2\sqrt{3}})=4-6(x+1)\Rightarrow x^3+3x^2+9x=-3\Rightarrow P=(x^3+3x^2+9x)^3=(-3)^3=-27$