Do a\(\ge\)-1
=>2a+3\(\ge\)0
=>(a-3)2(2a+3)\(\ge0\)
=> (a2-6a+9)(2a+3)\(\ge0\)
=>2a3+3a2-12a2-18a+18a+27\(\ge0\)
=> 2a3-9a2+27\(\ge0\)
=>2a3\(\ge\)9a2-27
TT=>2b3\(\ge9b^2-27\)
2c3\(\ge9c^2-27\)
=>2M\(\ge\)9(a2+b2+c2)-81=9.9-81=0
=>\(M\ge0\)
ta có:\(a\ge-1\Rightarrow a+1\ge0\)
mà\(\left(a-2\right)^2\ge0\)
\(\Rightarrow\)\(\left(a+1\right)\left(a-2\right)^2\ge0\)
\(\Leftrightarrow\)\(\left(a+1\right)\left(a^2-4a+4\right)\)\(\ge0\)
\(\Leftrightarrow a^3-4a^2+4a+a^2-4a+4\ge0\)
\(\Leftrightarrow a^3+4-3a^2\ge0\)
\(\Leftrightarrow a^3+4\ge3a^2\)
tương tự:\(b^3+4\ge3b^2;c^3+4\ge3c^2\)
\(\Rightarrow a^3+b^3+c^3+12\ge3\left(a^2+b^2+c^2\right)\)
mà\(a^2+b^2+c^2=9\)
\(\Rightarrow a^3+b^3+c^3\ge27-12=15\)
Dấu "=" xayr ra khi:
\(\left(a;b;c\right)=\left(-1;2;2\right);\left(2;2;-1\right);\left(2;-1;2\right)\)
ta co a^3 + 1 = (a+1) (a^2 -a +1)
= (a+1)( \(\frac{a^2}{4}\)- a+1 + \(\frac{3a^2}{4}\))
= ( a+1 ) (\(\frac{a^2}{4}\)- a+ 1) + \(\frac{3a^2}{4}\)( a+1)
= \(\frac{1}{4}\)( a+1 )(a-2)^2 + \(\frac{3a^2}{4}+\frac{3a^2}{4}\ge\frac{3a^3}{4}+\frac{3a^2}{4}\)
\(\Rightarrow a^3+1\ge\frac{3a^3}{4}+\frac{3a^2}{4}\)
\(\Leftrightarrow a^3\ge3a^2-4\)
CMTT \(b^2\ge3b^2-4\)
\(c^2\ge3c^2-4\)
\(\Rightarrow a^3+b^3+c^3\ge3\left(a^2+c^2+b^2\right)-12\)
\(\Rightarrow A\ge15\)
dau bang xay ra khi
(a+1)(a-2)^2=0
(b-1)(b-2)^2=0
(c-1)(c-2)^2=0
\(\Rightarrow\)(a;b;c)=(2;2;1) và các hoán vị
Vậy .........
