Track / Stage 5 / 157 of 400 #757 of 1964
Problem 757 AIME late Difficulty 5.2 Prove it
Three. (20 points) Let x , y , z ⩾ 0 , x + y + z = 3 x, y, z \geqslant 0, x+y+z=3 x , y , z ⩾ 0 , x + y + z = 3 x + y + z ⩾ x y + y z + z x \sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant x y+y z+z x x + y + z ⩾ x y + y z + z x
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Official solution Three, because 2 + ( x ) 3 = 1 + 1 + ( x ) 3 ⩾ 3 x 2+(\sqrt{x})^{3}=1+1+(\sqrt{x})^{3} \geqslant 3 \sqrt{x} 2 + ( x ) 3 = 1 + 1 + ( x ) 3 ⩾ 3 x 2 x + x 2 ⩾ 3 x 2 \sqrt{x}+x^{2} \geqslant 3 x 2 x + x 2 ⩾ 3 x 2 y + y 2 ⩾ 3 y , 2 z + z 2 ⩾ 3 z 2 \sqrt{y}+y^{2} \geqslant 3 y, 2 \sqrt{z}+z^{2} \geqslant 3 z 2 y + y 2 ⩾ 3 y , 2 z + z 2 ⩾ 3 z 2 ( x + y + z ) + x 2 + y 2 + z 2 2(\sqrt{x}+\sqrt{y}+\sqrt{z})+x^{2}+y^{2}+z^{2} 2 ( x + y + z ) + x 2 + y 2 + z 2 ⩾ 3 ( x + y + z ) = ( x + y + z ) 2 \geqslant 3(x+y+z)=(x+y+z)^{2} ⩾ 3 ( x + y + z ) = ( x + y + z ) 2 x + y + z = 3 x+y+z=3 x + y + z = 3 x + y + z ⩾ x y + y z + z x \sqrt{x}+\sqrt{y}+\sqrt{z} \geqslant x y+y z+z x x + y + z ⩾ x y + y z + z x
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Source: NuminaMath-1.5 ,
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