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

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 

Vì \(0\le a\le b\le c\le1\) nên:

\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)

Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)

Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)

Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)

Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)

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 \(a,b,c\in Z^+\)=> \(\frac{a}{a+b}>\frac{a}{a+b+c}\)\(\frac{b}{b+c}>\frac{b}{a+b+c}\)và \(\frac{c}{c+a}>\frac{c}{a+b+c}\)

\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

Giả sử \(a\ge b\ge c\)Ta có \(a,b,c\in Z^+\)và \(a\ge b\)\(\Rightarrow\)\(c+a\ge c+b\)\(\Rightarrow\frac{c}{c+a}\le\frac{c}{c+b}\Rightarrow\frac{b}{b+c}+\frac{c}{c+a}\le\frac{b}{b+c}+\frac{c}{c+b}=1\)

Do \(a,b,c\in Z^+\)\(\Rightarrow\frac{a}{a+b}< 1\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< 2\)

Vậy \(\frac{a}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\le2\)

Bài này lớp 7 là khó đấy \(0\le a\le b\le c\le1\Rightarrow\hept{\begin{cases}1-a\ge0\\1-b\ge0\end{cases}\Rightarrow\left(1-a\right)\left(1-b\right)\ge0}\)

\(\Leftrightarrow ab-a-b+1\ge0\Leftrightarrow ab+1\ge a+b\)(*)

Vì \(0\le a\le b\le c\le1\) nên \(\hept{\begin{cases}ab\ge0\\1\ge c\end{cases}\Rightarrow ab+1\ge c}\)Kết hợp với (*) ta được :

 \(2\left(ab+1\right)\ge a+b+c\) \(\Leftrightarrow\frac{1}{ab+1}\le\frac{2}{a+b+c}\Rightarrow\frac{c}{ab+1}\le\frac{2}{a+b+c}\)(1)

Chứng minh tương tự \(\hept{\begin{cases}\frac{a}{bc+1}\le\frac{2a}{a+b+c}\text{ }\left(2\right)\\\frac{b}{ac+1}\le\frac{2b}{a+b+c}\text{ }\left(3\right)\end{cases}}\)

Cộng vế với vế của (1);(2);(3) ta được :

\(\frac{a}{bc+1}+\frac{b}{ac+1}+\frac{c}{ab+1}\le\frac{2\left(a+b+c\right)}{a+b+c}=2\)(đpcm)