Câu cuối của đề HSG

N

nguyenbahiep1

[laTEX]Ta-chung-minh-dang-thuc: \sqrt{a^2+b^2} + \sqrt{c^2+d^2} \geq \sqrt{(a+c)^2+(b+d)^2} \\ \\ a^2+b^2+c^2+d^2 + 2\sqrt{a^2+b^2}.\sqrt{c^2+d^2} \geq c^2+d^2+a^2+b^2+2ac+2bd \\ \\ \sqrt{a^2+b^2}.\sqrt{c^2+d^2} \geq ac+bd \\ \\ (a^2+b^2)(c^2+d^2) \geq (ac)^2+(bd)^2+2abcd \\ \\ b^2c^2+a^2d^2 \geq 2abcd \Rightarrow dung -theo-cosi \\ \\ b^2c^2+a^2d^2 \geq 2 \sqrt{(abcd)^2} = 2abcd[/laTEX]


áp dụng vào bài trên ta có

[laTEX]\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}} + \sqrt{(\frac{1}{2}-x)^2+\frac{3}{4}} \geq \sqrt{(x+\frac{1}{2}+\frac{1}{2}-x)^2+(\frac{\sqrt{3}}{2} +\frac{\sqrt{3}}{2})^2} \\ \\ A \geq \sqrt{1+3} = 2 \\ \\ x = 0 [/laTEX]
 
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