\(\text{Ta có }:\left(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\right)^2\\ =x^2+y^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}+z^2+t^2\)
Áp dụng định lí bu-nhi-a-cốp-xki:
\(\Rightarrow2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}\ge2\sqrt{\left(xz+yt\right)^2}=2xz+2yt\\ \Rightarrow\left(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\right)^2\\ \ge x^2+y^2+2xz+2yt+z^2+t^2\\ =x^2+2xz+z^2+y^2+2yt+t^2\\ =\left(x+z\right)^2+\left(y+t\right)^2\\ \Rightarrow\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{x}{y}=\frac{z}{t}\)
Áp dụng BDT trên
\(\Rightarrow\sqrt{a^2+b^2-\sqrt{3}ab}+\sqrt{b^2+c^2-bc}\\ =\sqrt{\frac{3}{4}a^2-\sqrt{3}ab+b^2+\frac{1}{4}a^2}+\sqrt{b^2-bc+\frac{1}{4}c^2+\frac{3}{4}c^2}\\ =\sqrt{\left(\frac{\sqrt{3}}{2}a-b\right)^2+\frac{1}{4}a^2}+\sqrt{\left(b-\frac{1}{2}c\right)^2+\frac{3}{4}c^2}\\ \ge\sqrt{\left(\frac{\sqrt{3}}{2}a-b+b-\frac{1}{2}c\right)^2+\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}c\right)^2}\\ =\sqrt{\left(\frac{\sqrt{3}}{2}a-\frac{1}{2}c\right)^2+\left(\frac{1}{2}a+\frac{\sqrt{3}}{2}c\right)^2}\\ =\sqrt{\frac{3}{4}a^2-\frac{\sqrt{3}}{2}ac+\frac{1}{4}c^2+\frac{1}{4}a^2+\frac{\sqrt{3}}{2}ac+\frac{3}{4}c^2}\\ \\ =\sqrt{a^2+c^2}\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{\frac{\sqrt{3}}{2}a-b}{\frac{1}{2}a}=\frac{b-\frac{1}{2}c}{\frac{\sqrt{3}}{2}c}\)
\(\Leftrightarrow\frac{\sqrt{3}a-2b}{a}=\frac{2b-c}{\sqrt{3}c}\\ \Leftrightarrow\sqrt{3}c\left(\sqrt{3}a-2b\right)=a\left(2b-c\right)\\ \Leftrightarrow3ac-2\sqrt{3}bc=2ab-ac\\ \Leftrightarrow4ac-2\sqrt{3}bc-2ab=0\)
