Question

x,y,z属于正实数,求证:x4+y4+z4>=(x+y+z)xyz

Answer 1

x^4+y^4≥2x^2y^2

y^4+z^4≥2y^2z^2x0d

z^4+x^4≥2z^2x^2

三式相加得

x^4+y^4+z^4≥x^2y^2+y^2z^2+z^2x^2x0d

x^2y^2+y^2z^2≥2xy^2z

y^2z^2+z^2x^2≥2xyz^2x0d

z^2x^2+x^2y^2≥2x^2yzx0d

三式相加得

x^2y^2+y^2z^2+z^2x^2≥xy^2z+xyz^2+x^2yz=(x+y+z)xyz

∴x^4+y^4+z^4≥(x+y+z)xyz

均值不等式会吧

a+b≥2√ab(a>0,b>0)

∵(a-b)^2=(a+b)^2-4ab≥0

∴(a+b)^2≥4ab

∴a+b≥2√a