Question

Cho \(\sqrt{a}+\sqrt{b}+\sqrt{c}=1\)\

Tìm min P= \(\sqrt{2a^2+ab+2b^2}+\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2a^2}\)

Answer 1

Đề sai phải là \(\sqrt{2b^2+bc+2c^2}\)

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{5}{4}}\left(a+b\right)\)

CMTT, có: \(\sqrt{2b^2+bc+2c^2}\ge\sqrt{\frac{5}{4}}\left(b+c\right)\)

\(\sqrt{2c^2+ca+2a^2}\ge\sqrt{\frac{5}{4}}\left(c+a\right)\)

\(\Rightarrow P\ge\sqrt{5}\left(a+b+c\right)\ge\frac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\frac{\sqrt{5}}{3}\)

Dấu "=" xảy ra khi a=b=c=\(\frac{1}{9}\)