Ta có: \(\sqrt{a^2-ab+b^2}=\sqrt{\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2}\ge\sqrt{\dfrac{1}{4}\left(a+b\right)^2}=\dfrac{1}{2}\left(a+b\right)\)
Tương tự: \(\sqrt{b^2-bc+c^2}\ge\dfrac{1}{2}\left(b+c\right)\)
\(\sqrt{c^2-ca+a^2}\ge\dfrac{1}{2}\left(c+a\right)\)
\(P\ge\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(b+c\right)+\dfrac{1}{2}\left(c+a\right)=a+b+c=2019\)
Dấu "=" xảy ra <=> a = b = c = 673
Ta có: a2-ab+b2 = \(\dfrac{1}{4}\)(a+b)2+3(a-b)2\(\ge\)\(\dfrac{1}{4}\)(a+b)2
\(\Rightarrow\)\(\sqrt{a^2-ab+b^2}\ge\dfrac{1}{2}\)(a+b)
Dấu "=" xảy ra \(\Leftrightarrow\) a=b
CMTT ta có: \(\sqrt{b^2-bc+c^2}\)\(\ge\dfrac{1}{2}\)(b+c) \(\Leftrightarrow\) b=c
\(\sqrt{c^2-ca+c^2}\)\(\ge\dfrac{1}{2}\left(c+a\right)\Leftrightarrow\)c=a
\(\Rightarrow\) P\(\ge\) \(\dfrac{1}{2}2\left(a+b+c\right)\)= 2019
Vậy Pmin = 2019
Dấu "=" xảy ra\(\Leftrightarrow\)a=b=c=673
Cách 2:
Theo BĐT Cô si ta có: ab \(\le\) \(\dfrac{a^2+b^2}{2}\)
\(\Rightarrow\) -ab \(\le\) -\(\dfrac{a^2+b^2}{2}\)
\(\Rightarrow\) a2-ab+b2 \(\ge\) a2- \(\dfrac{a^2+b^2}{2}\)+b2 = \(\dfrac{a^2+b^2}{2}\)
Mặt khác ta có a2+b2 \(\ge\) \(\dfrac{\left(a+b\right)^2}{2}\)
\(\Rightarrow\) a2-ab+b2 \(\ge\) \(\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow\) \(\sqrt{a^2-ab+b^2}\)\(\ge\) \(\dfrac{|a+b|}{2}\)= \(\dfrac{a+b}{2}\)
CMTT: \(\sqrt{b^2-bc+c^2}\) \(\ge\)\(\dfrac{b+c}{2}\)
\(\sqrt{c^2-ca+a^2}\)\(\ge\)\(\dfrac{c+a}{2}\)
\(\Rightarrow\) P\(\ge\)\(\dfrac{2\left(a+b+c\right)}{2}\) = 2019
Vậy Pmin = 2019
Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=673
