证明(x^2+y^2)/2大于等于((x+y)/2) ^2
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证明(x^2+y^2)/2大于等于((x+y)/2) ^2
证明(x^2+y^2)/2大于等于((x+y)/2) ^2
证明(x^2+y^2)/2大于等于((x+y)/2) ^2
y=(x^2+y^2)/2-((x+y)/2) ^2
=(x^2+y^2)/2-(x+y)^2/4
=(2x^2+2y^2-x^2-2xy-y^2)/4
=[(x-y)/2]^2
>=0
所以(x^2+y^2)/2大于等于((x+y)/2) ^2
因为(x^2+y^2)/2-((x+y)/2) ^2=(2x^2+2y^2)/4-(x^2+2xy+y^2)/4=(x^2-2xy+y^2)/4=((x-y)/2) ^2>=0
所以(x^2+y^2)/2大于等于((x+y)/2) ^2
2x^2+2 y^2 - (x^2+y^2+2xy)
=(x-y)^2
(x-y)^2 大于等于 0
则有2x^2+2 y^2 大于等于 (x^2+y^2+2xy)
则 (x^2+y^2)/2大于等于 (x^2+y^2+2xy)/4=(x+y)^2/4=((x+y)/2) ^2
((x+y)/2) ^2 <=(x^2+y^2)/2
(X^2+2XY+Y^2)<=2(X^2=Y^2)
0<=X^2+Y^2-2XY
0<=(X-Y)^2
(x^2+y^2)/2-((x+y)/2) ^2
=(x^2+y^2)/2-[(x^2+y^2)/4+xy/2]
=(x^2+y^2)/2-xy/2
=(x^2+y^2-xy)/2
∵(x-y)^2≥0
x^2+y^2-2xy≥0
即x^2+y^2≥2xy
∴x^2+y^2≥xy
∴(x^2+y^2)/2-((x+y)/2) ^2=(x^2+y^2-xy)/2 ≥0
∴(x^2+y^2)/2大于等于((x+y)/2) ^2
∵x+y+z=1
∴x²+y²+z²+2xy+2yz+2xz=1
∵x²+y²+z²+x²+y²+y²+z²+x²+z²≥x²+y²+z²+2xy+2yz+2xz=1
∴x²+y²+z²≥1/3