Cho x > y > z. Chứng minh rằng biểu thức: A = x^4(y - z) + y^4(z - x) + z^4(x - y)
Cho x > y > z. Chứng minh rằng biểu thức:
A = x4(y − z) + y4(z − x) + z4(x − y) luôn luôn dương
Ta có:
A = x4(y − z) + y4(z − x) + z4(x − y)
= x4y − x4z + y4z − y4x + z4x − z4y
= xy(x3 − y3) − z(x4 − y4) + z4(x − y)
= xy(x − y)(x2 + xy + y2) − z(x − y)(x + y)(x2 + y2) + z4(x − y)
= (x − y)[xy(x2 + xy + y2) − z(x + y)(x2 + y2) + z4]
= (x − y)(x3y + x2y2 + xy3 − x3z − x2yz − xy2z − y3z + z4)
= (x − y)[x2y(x − z) + xy2(x − z) + y3(x − z) − z(x3 − z3)]
= (x − y)(x − z)[x2y + xy2 + y3 − z(x2 + xz + z2)]
= (x − y)(x − z)(x2y + xy2 + y3 − zx2 − xz2 − z3)
= (x − y)(x − z)[x2(y − z) + x(y2 − z2) + (y3 − z3)]
= (x − y)(x − z)[x2(y − z) + x(y − z)(y + z) + (y − z)(y2 + yz + z2)]
= (x − y)(x − z)(y − z)(x2 + y2 + z2 + xy + xz + yz)
Với x > y > z nên suy ra x − y > 0; x − z > 0; y − z > 0 (1)
Lại có: x2 + y2 + z2 + xy + xz + yz
\[ = \frac{1}{2}\left( {2{x^2} + 2{y^2} + 2{z^2} + 2xy + 2yz + 2xz} \right)\]
\[ = \frac{1}{2}\left[ {\left( {{x^2} + 2xy + {y^2}} \right) + \left( {{y^2} + 2yz + {z^2}} \right) + \left( {{x^2} + 2xz + {z^2}} \right)} \right]\]
\[ = \frac{1}{2}\left[ {{{\left( {x + y} \right)}^2} + {{\left( {y + z} \right)}^2} + {{\left( {x + z} \right)}^2}} \right] > 0,\;\forall x > y > z\] (2)
Từ (1) và (2) nên A luôn luôn dương.