Question

已知a,b,c为正实数,且abc=1,求证(1/a2)+(1/b2)+(1/c2)>=a+b+c

Answer 1

证明：由abc=1带入

有(1/a^2)+(1/b^2)+(1/c^2)=abc/a^2+abc/b^2+abc/c^2=bc/a+ac/b+ab/c

=1/2[(bc/a)+(ac/b)]+1/2[(bc/a)+(ab/c)]+1/2[(ac/b)+(ab/c)]

再根据基本不等式有

[(bc/a)+(ac/b)]>=2根号下[(bc/a)*(ac/b)]=2c

[(bc/a)+(ab/c)]>=2根号下[(bc/a)*(ab/c)]=2b

[(ac/b)+(ab/c)]>=2根号下[(ac/b)*(ab/c)]=2a

再把上面的3个式子相加得到

(1/a2)+(1/b2)+(1/c2)>=a+b+c