$\sqrt{x^2 + (1+\sqrt{3})x + 2} + \sqrt{x^2 + (1-\sqrt{3})x + 2}$ \leq $3\sqrt{2} - \sqrt{x^2 - 2x + 2}$
\Leftrightarrow $\sqrt{x^2 + (1+\sqrt{3})x + 2} + \sqrt{x^2 + (1-\sqrt{3})x + 2} + \sqrt{x^2 - 2x + 2}$ \leq $3\sqrt{2} $
\Leftrightarrow $\sqrt{2x^2 + 2(1+\sqrt{3})x + 4} + \sqrt{2x^2 + 2(1-\sqrt{3})x + 4} + \sqrt{2x^2 - 4x + 4}$ \leq 6
\Leftrightarrow $\sqrt{(1+x)^2 + (\sqrt{3} + x)^2} + \sqrt{(1+x)^2 + (\sqrt{3} - x)^2} + \sqrt{2x^2 - 4x + 4}$ \leq 6 (1)
Đặt: $\overrightarrow{u}$ = $(1+x;\sqrt{3} + x)$ ; $\overrightarrow{v}$ = $(1+x; \sqrt{3} -x).$
Ta có: $|\overrightarrow{u}|$ + $|\overrightarrow{v}|$ \geq |$\overrightarrow{u} + \overrightarrow{v}$|
\Leftrightarrow $\sqrt{(1+x)^2 + (\sqrt{3} + x)^2} + \sqrt{(1+x)^2 + (\sqrt{3} - x)^2}$ \geq $\sqrt{(2+2x)^2 + 12}$
VT(1) \geq $\sqrt{(2+2x)^2 + 12} + \sqrt{2x^2 - 4x + 4} = \sqrt{4x^2 + 8x + 16} + \sqrt{2x^2 - 4x + 4} = f(x).$
Khảo sát hàm f(x) \Rightarrow $f_{min} = f(0) = 6$ \Rightarrow VT(1) \geq 6.
Vậy BPT có nghiệm duy nhất là x = 0.
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