Question

设abc为正整数,求证1/a^3+1/b^3+1/c^3+abc>=2倍根号三

Answer 1

证①∵ a,b,c为正实数,

∴ 1/a^3+1/b^3+1/c^3+abc=1/a^3+1/b^3+1/c^3+abc/3+abc/3+abc/3

≥6*6次√（1/a^3+1/b^3+1/c^3+abc/3+abc/3+abc/3）=6*6次√（1/3^3)

=2*6次√(3^6/3^3)≥=√2*6次√(3^3)=2√3

②∵1/a^3+1/b^3+1/c^3≥3*3次√[1/(a^3b^3c^3)]

∴1/a^3+1/b^3+1/c^3≥3/abc

又∵ 1/a^3+1/b^3+1/c^3+abc≥2√(3/abc*abc)=2√3.