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Ta có :

\(\sqrt{a^2+ab+b^2}\ge\dfrac{\sqrt{3}}{2}\left(a+b\right)\) với mọi a , b

Dấu = xảy ra khi và chỉ khi \(a=b>0\)

Thật vậy :

Với \(a+b< 0\) thì \(\dfrac{\sqrt{3}}{2}\)luôn thỏa mãn 

Với \(a+b\ge0\) thì bình phương

Hai vế của \(\dfrac{\sqrt{3}}{2}\) ta có :

\(a^2+ab+b^2\ge\dfrac{3}{4}\left(a+b\right)^2\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) luôn đúng

\(\Rightarrow\dfrac{\sqrt{3}}{2}\left(a+b\right)\) luôn đúng 

Tương tự ta có :

\(\sqrt{b^2+bc+c^2}\ge\dfrac{\sqrt{3}}{2}\left(b+c\right)\)

\(\sqrt{c^2+ca+a^2}\ge\dfrac{\sqrt{3}}{2}\left(c+a\right)\)

\(\Rightarrow p\ge\sqrt{3}\left(a+b+c\right)\)

\(=9\sqrt{3}\)

Mặt khác a = b = c = 3 thì \(p=9\sqrt{3}\)

Vậy minP \(=9\sqrt{3}\)

we have \(ab\le\dfrac{\left(a+b\right)^2}{4}\)

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

 similar \(\sqrt{b^2+bc+c^2}\ge\dfrac{\sqrt{3}\left(b+c\right)}{2}\)

​ ​\(\sqrt{c^2+ac+a^2}\ge\dfrac{\sqrt{3}\left(a+c\right)}{2}\)

\(p\ge\dfrac{\sqrt{3}\left(a+b\right)}{2}+\dfrac{\sqrt{3}\left(a+c\right)}{2}+\dfrac{\sqrt{3}\left(b+c\right)}{2}=\dfrac{\sqrt{3}}{2}.2\left(a+b+c\right)=\sqrt{3}.9=9\sqrt{3}\)

so max \(p=9\sqrt{3}\)

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