Áp dụng bất đẳng thức $Minkowski$ ta được:
$\sqrt{1 + a^2} +\sqrt{1 +b^2} +\sqrt{1+b^2}\geq \sqrt{(1+1+1)^2 +(a+b+b)^2}=\sqrt{9 + (a+2b)^2}$
Tương tự, ta được:
$\sqrt{1+b^2} +\sqrt{1+c^2} +\sqrt{1+c^2}\geq \sqrt{9 + (b+2c)^2}$
$\sqrt{1+c^2} +\sqrt{1+a^2}+\sqrt{1+a^2}\geq \sqrt{9 + (c+2a)^2}$
Cộng vế theo vế ta được:
$3(\sqrt{1 + a^2} +\sqrt{1 +b^2} +\sqrt{1+c^2})\geq \sqrt{9 + (a+2b)^2} +\sqrt{9+(b+2c)^2} + \sqrt{9+(c+2a)^2}$
$\to \sqrt{1 + a^2} +\sqrt{1 +b^2} +\sqrt{1+c^2} \geq \dfrac13\sqrt{9 + (a+2b)^2} +\dfrac13\sqrt{9+(b+2c)^2} + \dfrac13\sqrt{9+(c+2a)^2}$
$\to \sqrt{1 + a^2} +\sqrt{1 +b^2} +\sqrt{1+c^2}\geq \sqrt{1+ \dfrac{(a+2b)^2}{9}} +\sqrt{1+\dfrac{(b+2c)^2}{9}} + \sqrt{1+\dfrac{(c+2a)^2}{9}}$
$\to \sqrt{1 + a^2} +\sqrt{1 +b^2} +\sqrt{1+c^2}\geq \sqrt{1+ \left(\dfrac{a+2b}{3}\right)^2} +\sqrt{1+\left(\dfrac{b+2c}{3}\right)^2} + \sqrt{1+\left(\dfrac{c+2a}{3}\right)^2}$
Dấu $=$ xảy ra $\Leftrightarrow a = b = c$
