Question

C\(\)ho a+b+c=6. tìm giá trị nhỏ nhất của S=\(^{a^2+b^2+c^2}\)

Answer 1

Mới làm, cô chưa chữa nên khum chắc :33

Có

\(\left(a-b\right)^2+\left(c-a\right)^2+\left(b-a\right)^2\)

\(=a^2-2ab+b^2+c^2-2ac+a^2+b^2-2ab+a^2\)

\(=2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ac\right)\)

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

Mà \(\left(a-b\right)^2+\left(c-a\right)^2+\left(b-a\right)^2\ge0\forall a;b;c\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)-\left(a+b+c\right)^2\ge0\forall a;b;c\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\forall a;b;c\)

\(\Rightarrow3S\ge\left(a+b+c\right)^2\forall a;b;c\)

Mà \(a+b+c=6\)

\(\Rightarrow3S\ge36\) \(\forall a;b;c\)

\(\Rightarrow S\ge12\)\(\forall a;b;c\)

Dấu "=" xảy ra \(\Leftrightarrow\) \(\left(a-b\right)^2=\left(c-a\right)^2=\left(b-a\right)^2\)

\(\Leftrightarrow a=b=c\)

Answer 2

Cách khác:

Áp dụng BĐT Cô-si:

\(S=a^2+b^2+c^2=\left(a^2+4\right)+\left(b^2+4\right)+\left(c^2+4\right)-12\)

\(\ge4a+4b+4c-12=4\left(a+b+c\right)-12=12\)

\(MinS=12\Leftrightarrow a=b=c=2\)