\(\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)

\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)

\(\Rightarrow2\left(ab+bc+ca\right)\ge4\)

\(\Rightarrow-2\left(ab+bc+ca\right)\le-4\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\)(Đpcm)

Dấu = khi \(\hept{\begin{cases}\left(2-a\right)\left(2-b\right)\left(2-c\right)=0\\abc=0\\a+b+c=3\end{cases}}\)

\(\Rightarrow\left(a;b;c\right)=\left(2;1;0\right)\)và hoán vị.

a = 2 ( t/m )

b = 1 ( t/m )

c = 0 ( t/m )

vậy \(a^2+b^2+c^2\le5\)

Vì \(0\le a\le2;0\le b\le2;0\le c\le2\Rightarrow\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)\(\Leftrightarrow8-4\left(a+b+c\right)+2\left(ab+bc+ca\right)-abc\ge0\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\left(a+b+c\right)-8+abc\ge4\)\(\Leftrightarrow2\left(ab+bc+ca\right)\ge12-8+abc\ge4\)

\(\Rightarrow\)\(2\left(ab+bc+ca\right)\ge4\)

\(\Leftrightarrow-2\left(ab+bc+ca\right)\le-4\)

Ta có :

\(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\Rightarrowđpcm\)Đẳng thức xảy ra khi

\(\left(2-a\right)\left(2-b\right)\left(2-c\right)=0\)

\(\left[{}\begin{matrix}2-a=0\\2-b=0\\2-c=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)

\(-1\le a\le2\Rightarrow\hept{\begin{cases}a+1\ge0\\a-2\le0\end{cases}\Rightarrow\left(a+1\right)\left(a-2\right)\le0}\)

Tương tự \(\left(b+1\right)\left(b-2\right)\le0,\left(c+1\right)\left(c-2\right)\le0\)

=> (a+1)(a-2)+(b+1)(b-2)+(c+1)(c-2)\(\le\)0 => a²+b²+c²-(a+b+c)-6\(\le\)0 

=>a²+b²+c² \(\le\)6 

Dấu "=" xảy ra <=> (a+1)(  a-2)=0, (b+1)(b-2)=0, (c+1)(c-2)=0 , a+b+c=0 <=> a=2, b=c=-1 và các hoán vị 

Ta có \(\left(a+2\right)\left(b+2\right)\left(c+2\right)+\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)

\(\Leftrightarrow4\left(ab+bc+ca\right)+16\ge0\)

\(\Leftrightarrow ab+bc+ca\ge-4\).

Lại có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=0\).

Do đó \(\left(ab+bc+ca\right)^2\le16\).

Mặt khác do \(a+b+c=0\) nên dễ dàng chứng minh được \(2\left(a^4+b^4+c^4\right)=\left(ab+bc+ca\right)^2\) (Bạn xem ở đây).

Do đó \(a^4+b^4+c^4\le32\) (đpcm).

Do \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\a+b+c=5\end{matrix}\right.\) \(\Rightarrow c\le4\)

\(\Rightarrow2\le c\le4\Rightarrow\left(c-2\right)\left(c-4\right)\le0\Rightarrow c^2\le6c-8\)

\(0\le a\le1< 6\Rightarrow a\left(a-6\right)\le0\Rightarrow a^2\le6a\)

\(1\le b\le2< 5\Rightarrow\left(b-1\right)\left(b-5\right)\le0\Rightarrow b^2\le6b-5\)

Cộng vế:

\(a^2+b^2+c^2\le6\left(a+b+c\right)-13=17\)

\(A_{max}=17\) khi \(\left(a;b;c\right)=\left(0;1;4\right)\)

từ giả thuyết suy ra : abc >0

có 2>a,c,b ->> (2-a)(2-b)(2-c)\(\ge\)0

\(\Leftrightarrow\)8+2(ab+ac+bc) -4(a+b+c)-abc \(\ge\)0

\(\Leftrightarrow\)8+2(ab+ac+bc)-4.3-abc \(\ge\)0

\(\Leftrightarrow\)2(ab+ac+bc) \(\ge\)4+abc \(\ge\)4 (1)

Cộng a²+b²+c² vào (1)

2(ab+ac+bc)+a²+b²+c²\(\ge\)4+a²+b²+c²

(a+b+c)²-4\(\ge\)a²+b²+c²

^{thay a+b+c=3 vào}

9-4^\(\ge\)a²+b²+c²

^5 \(\ge\)a²+b²+c²

a²+b²+c² \(\le\)5

cauhc lop may

Từ a+b+c=6 \(\Rightarrow\)a+b=6-c

Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9

                               \(\Leftrightarrow\)ab=9-c(a+b)

           Mà a+b=6-c (cmt)

                                \(\Rightarrow\)ab=9-c(6-c)

                                \(\Rightarrow\)ab=9-6c+c²

Ta có: (b-a)²\(\ge\)0 \(\forall\)b, c

  \(\Rightarrow\)b²+a²-2ab\(\ge\)0

  \(\Rightarrow\)(b+a)²-4ab\(\ge\)0

  \(\Rightarrow\)(a+b)²\(\ge\)4ab

Mà a+b=6-c (cmt)

         ab= 9-6c+c² (cmt)

  \(\Rightarrow\)(6-c)²\(\ge\)4(9-6c+c²)

  \(\Rightarrow\)36+c²-12c\(\ge\)36-24c+4c²

  \(\Rightarrow\)36+c²-12c-36+24c-4c²\(\ge\)0

  \(\Rightarrow\)-3c²+12c\(\ge\)0

  \(\Rightarrow\)3c²-12c\(\le\)0

  \(\Rightarrow\)3c(c-4)\(\le\)0

  \(\Rightarrow\)c(c-4)\(\le\)0

\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)

*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)

*