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

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

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

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

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

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

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

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)\le9-4=5\)(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 \(a^2+b^2+c^2\le5\)

từ gt \(\Rightarrow\)abc>0  => (2-a)(2-b)(2-c)>0 => 
8+2(ab+bc+ca)−4(a+b+c)−abc≥0 => 2(ab+bc+ca) \(\ge\)4 + abc \(\ge\)4
=> (a+b+c)^2≥4+a2+b2+c2 => a^2+b^2+c^2 \(\le\) 5
 

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel và bất đẳng thức AM-GM  ta có :

\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}=\frac{\left(a^3+b^3+c^3\right)\left(a^3+b^3+c^3\right)}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)( đpcm )

Đẳng thức xảy ra <=> a=b=c

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

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

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