`(x^2)/x+(y^2)/y+(z^2)/z>=[(x+y+c)^2]/(x+y+z)`
`+)(a^2)/x+(b^2)/y>=[(a+b)^2]/(x+y)`
`->(a^2y+b^2x)/(xy)>=[(a+b)^2]/(x+y)`
`->a^2xy+a^2y^2+b^2x^2+b^2xy>=xy(a^2+2ab+b^2)`
`->a^2xy+a^2y^2+b^2x^2+b^2xy>=a^2xy+2axby+b^2xy`
`->a^2y^2-2aybx+b^2x^2>=0`
`->(ay-bx)^2>=0`
dấu = xảy ra khi `ax=by \harr a/y=b/x`
Áp dụng BĐT trên
`->(x^2)/a+(y^2)/b+(z^2)/c>=(x+y)^2/(a+b)+(z^2)/c`
`->(x^2)/a+(y^2)/b+(z^2)/c>=[(x+y+z)^2]/(a+b+c)`
`(ax+by+cz)^2<=(a^2+b^2+c^2)(x^2+y^2+z^2)`
`->a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^z^2>=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz`
`->a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz+c^2y^2>=0`
`->(ay-bx)^2+(az-cx)^2+(bz-cy)^2>==0` luôn đúng
