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Ta có: $\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{(a+c)^2+(b+d)^2},\forall a,b,c,d\in\mathbb{R}$.
Áp dụng, ta có:
$ \sqrt{a^2-\sqrt2ab+b^2}+\sqrt{b^2-\sqrt3bc+c^2}$
$=\sqrt{(b-\frac{a}{\sqrt2})^2+(\frac{a}{\sqrt2})^2}+\sqrt{(\frac{\sqrt3c}{2}-b)^2+(\frac{c}{2})^2}$
$\geq \sqrt{( \frac{\sqrt3c}{2} -\frac{a}{\sqrt2})^2+( \frac{a}{\sqrt2}+ \frac{c}{2})^2}$
$= \sqrt{a^2-\sqrt{ 2-\sqrt{ 3} }ac+c^2 } $
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