Áp dụng bất đẳng thức \(AM-GM\) cho 2 số dương ta có:

\(\left\{{}\begin{matrix}\dfrac{a+b}{2}\ge\sqrt{ab}\\\dfrac{b+c}{2}\ge\sqrt{bc}\\\dfrac{a+c}{2}\ge\sqrt{ac}\end{matrix}\right.\)

Cộng theo 3 vế ta có:

\(\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{a+c}{2}\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

\(\Rightarrow\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}b+\dfrac{1}{2}c+\dfrac{1}{2}a+\dfrac{1}{2}c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

\(\Rightarrow a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\left(đpcm\right)\)

\(a=b=c\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=2ab\\b^2+c^2=2bc\\a^2+c^2=2ac\end{matrix}\right.\)

Cộng theo 3 vế ta có:

\(a^2+b^2+b^2+c^2+a^2+c^2=2ab+2bc+2ac\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)

Ngược lại,khi \(a\ne b\ne c\) thì \(\left\{{}\begin{matrix}a^2+b^2>2ab\\b^2+c^2>2bc\\a^2+c^2>2ac\end{matrix}\right.\) ta có thể dễ dàng cm được \(a^2+b^2+c^2>ab+bc+ac\)

\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)

Tương tự: \(\sqrt{b+ac}\ge b+\sqrt{ac}\) ; \(\sqrt{c+ab}\ge c+\sqrt{ab}\)

\(\Rightarrow VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}\)

\(\Rightarrow VT\ge a+b+c=1\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Biến đổi tương đương:

\(2a+2b+2c\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\)

\(\Leftrightarrow a-2\sqrt{ab}+b+b-2\sqrt{bc}+c+c-2\sqrt{ca}+a\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\) (luôn đúng)

Dấu "=" xảy ra khi \(a=b=c\)