因?yàn)閍,b,c∈R+
所以：
(bc/2a)+(ac/2b)≥2√[(bc/2a)(ac/2b)]=2√(abc^2/4ab)=c
(bc/2a)+(ab/2c)≥2√[(bc/2a)(ab/2c)]=2√(acb^2/4ac)=b
(ac/2b)+(ab/2c)≥2√[(ac/2b)(ab/2c)]=2√(bca^2/4bc)=a
三式相加即得:
(bc/a)+(ac/b)+(ab/c)≥a+b+c
2)a+b+c=1
由基本不等式：(a+b+c)/3<=根號(hào)[(a^2+b^2+c^2)/3],等號(hào)當(dāng)且僅當(dāng)a=b=c時(shí)成立
所以根號(hào)a+根號(hào)b+根號(hào)c<=3根號(hào)[(a+b+c)/3]=根號(hào)3
等號(hào)當(dāng)且僅當(dāng)a=b=c=1/3時(shí)成立