Question

设a，b，c为正实数，求证1／a3+1／b3+1／c3+abc≥2√3

Answer 1

a，b，c为正实芹巧数，
∴春蔽1／a^3+1／b^3+1／c^3>=3/(abc),
∴嫌森键1／a^3+1／b^3+1／c^3+abc>=3/(abc)+abc>=2√3.

Answer 2

(ax2+by2)*1
=(ax2+by2)*(a+b)
=a2x2+b2y2+abx2+aby2
>=a2x2+b2y2+2abxy
=(ax+by)2
展开式中后逗段旅两项用均值不等燃滑式就可以山凳了

Answer 3

设u=1/a,v=1/b,t=1/c,那么上式改告铅写为u^3+v^3+t^3+1/uvt≥2√3,只让圆要证u^3+v^3+t^3≥3uvt就行了，假设袜滑好u≥t,v≥t再设u=t+x,v=t+y代入不等式中上述不等式显然成立