Question

Cho a+b+c=2. Tìm min \(K=a^2+b^2+c^2\)

Answer 1

Áp dụng bất đẳng thức Caushy dạng engel, ta có:

\(a^2+b^2+c^2=\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{c^2}{1}\ge\dfrac{\left(a+b+c\right)^2}{1+1+1}\)\(=\dfrac{4}{3}\)\(\Leftrightarrow a^2=b^2=c^2=\dfrac{4}{9}\Leftrightarrow a=b=c=\dfrac{2}{3}\)

Answer 2

Áp dụng BĐT Bunyakovsky,ta có:

\(\left(a+b+c\right)^2\le\left(1+1+1\right)\left(a^2+b^2+c^2\right)\).

\(a^2+b^2+c^2\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{4}{3}\)

\(Min_K=\dfrac{4}{3}\Leftrightarrow a=b=c=\dfrac{2}{3}\)