Track / Stage 6 / 114 of 400 #1114 of 1964
Problem 1114 National Olympiad, first round Difficulty 6.1 Prove it
Example 1.16 (Han Jingjun) x , y ∈ R x, y \in \mathbf{R} x , y ∈ R x 2 + y 2 = 1 x^{2}+y^{2}=1 x 2 + y 2 = 1 1 − x + 1 − 1 2 x − 3 2 y ⩾ 2 2 \sqrt{1-x}+\sqrt{1-\frac{1}{2} x-\frac{\sqrt{3}}{2} y} \geqslant \frac{\sqrt{2}}{2} 1 − x + 1 − 2 1 x − 2 3 y ⩾ 2 2
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Official solution Prove that after completing the square inside the square root, it is equivalent to2 2 [ ( 1 − x ) 2 + y 2 + ( x − 1 2 ) 2 + ( y − 3 2 ) 2 ] ⩾ 2 2 \frac{\sqrt{2}}{2}\left[\sqrt{(1-x)^{2}+y^{2}}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2}\right)^{2}}\right] \geqslant \frac{\sqrt{2}}{2} 2 2 ( 1 − x ) 2 + y 2 + ( x − 2 1 ) 2 + ( y − 2 3 ) 2 ⩾ 2 2
By Minkowski's inequality, we have( 1 − x ) 2 + y 2 + ( x − 1 2 ) 2 + ( y − 3 2 ) 2 ⩾ ( 1 − 1 2 ) 2 + ( 3 2 ) 2 = 1 \sqrt{(1-x)^{2}+y^{2}}+\sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{\sqrt{3}}{2}\right)^{2}} \geqslant \sqrt{\left(1-\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=1 ( 1 − x ) 2 + y 2 + ( x − 2 1 ) 2 + ( y − 2 3 ) 2 ⩾ ( 1 − 2 1 ) 2 + ( 2 3 ) 2 = 1
Hence, the proof is complete.
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Source: NuminaMath-1.5 ,
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