Question

[Anh Phạm Xuân]

Cho a,b,c >0 , a+b+c=2019 Tìm Min

\(P=\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)

Answer 1

[Eren]

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

Answer 2

[Tinh Lãm]

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 P_min = 2019

Dấu "=" xảy ra\(\Leftrightarrow\)a=b=c=673