Question

Cho a, b, c > 0 thỏa mãn \(\sqrt{a}+\sqrt{b}+\sqrt{c}=1\). Tìm \(P_{min}=\sqrt{2a^2+ab+2b^2}+\sqrt{2b^2+bc+2c^2}+\sqrt{2c^2+ca+2a^2}\)

Answer 1

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

Cmtt ta có: \(\sqrt{2b^2+bc+2c^2}\ge\sqrt{\dfrac{5}{4}}\left(b+c\right)\)

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

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

Dấu "=" xảy ra <=> a = b = c = \(\dfrac{1}{9}\)