Question

cho a,b,c>0 và a+b+c=1.Tìm GTNN:
\(M=\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)

Answer 1

\(\sqrt{2}M=\sqrt{\left(a-b\right)^2+\left(a^2+b^2\right)}+\sqrt{\left(b-c\right)^2+\left(b^2+c^2\right)}+\sqrt{\left(c-a\right)^2+\left(c^2+a^2\right)}\ge\sqrt{2ab}+\sqrt{2bc}+\sqrt{2ca}\)\(\Leftrightarrow M\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Dấu bằng xảy ra khi và chỉ khi a = b, b = c, c = a \(\Leftrightarrow\)a = b = c = \(\frac{1}{3}\)(vì a + b + c = 1).

Suy ra : \(M\ge\sqrt{\frac{1}{3}.\frac{1}{3}}+\sqrt{\frac{1}{3}.\frac{1}{3}}+\sqrt{\frac{1}{3}.\frac{1}{3}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)

Vậy GTNN của M là 1 khi a = b = c = \(\frac{1}{3}\)