a,b,c大于0，求证下面的不等式
\u5982\u679ca\u5927\u4e8eb,c\u5c0f\u4e8e0,\u90a3\u4e48\u4e0b\u5217\u4e0d\u7b49\u5f0f\u6210\u7acb\u7684\u662fA\u6210\u7acb
B\u4e2d\u7684c-a\u4e0ec-b\u76f8\u5f53\u4e8ea\u548cb\u5206\u522b\u4e58\u4e86\u4e00\u4e2a-1\u540e\u518d\u52a0c\uff0c\u5e94\u8be5\u6539\u53d8\u4e0d\u7b49\u53f7\u7684\u65b9\u5411\u3002
C\u4e2d\u56e0\u4e3ac\u5c0f\u4e8e0\uff0c\u6240\u4ee5\u4e0d\u7b49\u53f7\u65b9\u5411\u4e5f\u5e94\u8be5\u6539\u53d8\u3002
\u5de6\u8fb9=\u03a3a^4/(ab+ac)
>=(\u03a3a^2)^2/\u03a3(ab+ac) \uff08\u67ef\u897f\u4e0d\u7b49\u5f0f\u63a8\u8bba\uff1a\u03a3x[n]^2/y[n]>=(\u03a3x[n])^2/\u03a3y[n]\uff0c\u8fd9\u91ccx[n]=a^2\u4ec0\u4e48\u7684\uff0cy[n]=ab+ac\u4ec0\u4e48\u7684\u3002\u63a8\u8bba\u7684\u8bc1\u660e\u5c31\u662f\u628a\u53f3\u8fb9\u7684\u5206\u6bcd\u4e58\u5230\u5de6\u8fb9\u53bb\uff0c\u5c31\u662f\u67ef\u897f\u4e0d\u7b49\u5f0f\u4e86\u3002\uff09
=1/2*(\u03a3a^2)^2/\u03a3ab
>=1/2*(\u03a3ab)^2/\u03a3ab \uff08\u56e0\u4e3a\u03a3a^2>=\u03a3ab\uff0c\u8fd9\u4e2a\u628a\u53f3\u8fb9\u79fb\u5230\u5de6\u8fb9\uff0c\u914d\u65b9\u5c31\u5f97\u5230\u03a3(a-b)^2>=0\u4e86\uff09
=1/2*\u03a3ab
先证 (a^2+2)(b^2+2)(c^2+2) ≥3(a+b+c)^2.
(一)配方法
(a^2+2)(b^2+2)(c^2+2)-3(a+b+c)^2
=(a^2+2)(bc-1)^2+(a^2+2)(b-c)^2/2+3(2-ab-ac)^2/2≥0
(二)局部不等式
因为
(b^2+2)(c^2+2)-3[2+(b+c)^2]/2.
=(bc-1)^2+3(b-c)^2/2>=0
所以
(b^2+2)(c^2+2)>=3[2+(b+c)^2]/2.
再由柯西不等式
(a^2+2)(b^2+2)(c^2+2)>=3(a^2+2)*[1+(b+c)^2/2]
>=3(a+b+c)^2
∵(b-c)^2+2(bc-1)^2≥0,
∴(b^2+2)(c^2+2)≥3[1+(b+c)^2/2]
由Cauchy不等式知
(a^2+2)[1+(b+c)^2/2]≥(a+b+c)^2.
∴(a^2+2)(b^2+2)(c^2+2)
≥3(a+b+c)^2
≥9(ab+bc+ca).
p.s (a+b+c)^2-3(ab+bc+ca)=a^2+b^2+c^2-(ab+bc+ca)
=0.5[(a-b)^2+(b-c)^2+(c-a)^2] ≥0
=> (a+b+c)^2≥3(ab+bc+ca)
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