Verified answer

nếu c/m bđt trên thì làm cách này nhanh hơn nè

có (x-y)^2 >= 0

=> x^2+y^2>=2xy

tương tự : x^2+z^2>=2xz

y^2+z^2>=2yz

=> 2(x^2+y^2+z^2)>=2(xy+xz+yz)

=> x^2+y^2+z^2>=xy+yz+xz

=> (xy + yz + xz)^2 <= (x^2 + y^2 + z^2)(x^2 + y^2 + z^2)

dấu bằng xảy ra <=> x=y=z

*** (x^2 + y^2 + z^2)(x^2 + y^2 + z^2)

= x^4 + x^2y^2 + x^2z^2 + x^2y^2 + y^4 +y^2z^2 + x^2z^2 +z^2y^2 + z^4

= x^4 + y^4 + z^4 + 2 x^2y^2 + 2x^2z^2 + 2y^2z^2

*** (xy + yz + xz )^2

= x^2y^2 + y^2z^2 + z^2x^2 + 2 xz^2y + 2xzy^2 + 2yzx^2

****(xy + yz + xz)^2 <= (x^2 + y^2 + z^2)(x^2 + y^2 + z^2)

<=> x^2y^2 + y^2z^2 + z^2x^2 + 2 xz^2y + 2xzy^2 + 2yzx^2 <= x^4 + y^4 + z^4 + 2 x^2y^2 + 2x^2z^2 + 2y^2z^2

<=> 0 <= x^4 + y^4 + z^4 + x^2y^2 + x^2z^2 + y^2z^2 - z^2x^2 - 2 xz^2y - 2xzy^2 - 2yzx^2

<=> 0<= (x^4 - 2yzx^2 + y^2z^2 ) + ( y^4 - 2xzy^2 + x^2z^2 ) + (z^4 - 2xz^2y + x^2y^2 )

<=> 0 <= (x^2 -yz)^2 + ( y^2 -xz)^2 + (z^2-xy)^2 ( đúng)

ĐTXR <=> x/y = z/x = y/z