高中数学不等式竞赛题一道
\u4e00\u9053\u4e0d\u7b49\u5f0f\u7684\u9ad8\u4e2d\u6570\u5b66\u7ade\u8d5b\u9898\u7b2c\u4e00\u95ee
\u9996\u5148\uff0ca b c\u4e3a\u6b63\u6570\uff0c\u5219\u80af\u5b9a1/\uff08a+1\uff09\u30001/\uff08b+1\uff09\u30001/\uff08b+1\uff09\u90fd\u5c0f\u4e8e1\u7684\uff0e\u4e09\u4e2a\u4e4b\u548c\u5c31\u5c0f\u4e8e
3,\u6240\u4ee5\u4e0d\u7b49\u5f0f\u524d\u9762\u4e00\u90e8\u5206\u6210\u7acb\uff0e
\u4e0d\u7b49\u5f0f\u540e\u9762\u4e00\u90e8\u5206\u8bc1\u6cd5\uff0e
1/\uff08a+1\uff09+1/\uff08b+1\uff09+1/\uff08b+1\uff09\u22653\u500d3\u6b21\u6839\u53f7{1/[(a+1\uff09(b+1)(c+1)]}
\u56e0\u4e3a
(a+1\uff09(b+1)(c+1)\u2264[(a+1)+(b+1)+(c+1)]^3/27=(a+b+c+3)^3/27\u2264(3+3)^3/27=8
\u52191/[(a+1\uff09(b+1)(c+1)]\u22651/8
\u6240\u4ee5
1/\uff08a+1\uff09+1/\uff08b+1\uff09+1/\uff08b+1\uff09\u22653\u500d3\u6b21\u6839\u53f7{1/[(a+1\uff09(b+1)(c+1)]}\u22653\u500d3\u6b21\u6839\u53f7\uff081/8\uff09\uff1d3/2,\u5f53\u4e14\u4ec5\u5f53a=b=c=1\u65f6\u53d6\u7b49\u53f7
\u7b2c\u4e8c\u95ee
\uff08a+1\uff09/(a(a+2))=0.5[(2a+2)/(a(a+2)]=0.5[(a+(a+2))/(a(a+2)]=0.5[1/a+1/(a+2)]
\u6240\u4ee5\u539f\u5f0f\uff1d0.5[\uff081/a+1/b+1/c)+1/(a+2)+(1/(a+2)+1/(b+2)+1/(c+2))]
\u5229\u7528\u7b2c\u4e00\u95ee\u8bc1\u660e1/[(a+1\uff09+1/(b+1)+1/(c+1)\u22653/2\u7684\u65b9\u6cd5\u540c\u6837\u53ef\u4ee5\u8bc1\u5f97
1/a+1/b+1/c\u22653
1/(a+2)+1/(b+2)+1/(c+2)\u22651
\u6240\u4ee5\u539f\u5f0f\uff1d0.5[\uff081/a+1/b+1/c)+1/(a+2)+(1/(a+2)+1/(b+2)+1/(c+2))]\u22650.5\uff083+1\uff09\uff1d2
\u4e0d\u7b49\u5f0f\u6210\u7acb
\u5176\u5b9e\u8fd9\u9898\u76ee\u5f88\u953b\u70bc\u601d\u7ef4\u7684,\u4e0b\u9762\u662f\u6211\u7684\u89e3\u7b54,\u5927\u5bb6\u770b\u770b\u5bf9\u4e0d\u5bf9\u3002(\u770b\u56fe\u7247,\u6587\u5b57\u662flatex\u4ee3\u7801)
\u7531\u4e8e\u5bf9\u4e8e\u4efb\u610f$x,y,z \ge 0$,\u6709$(x+y+z)^2 \ge 3(xy+yz+zx)$.
\u628a$x=bc,y=ca,z=ab$\u4ee3\u5165\u5f97\u5230,$(bc+ca+ab)^2 \ge 3abc(a+b+c)=9abc$,\u6240\u4ee5$ab+bc+ca \ge 3\sqrt{abc}$
\u6240\u4ee5\u7531\u5e73\u5747\u503c\u4e0d\u7b49\u5f0f\u5f97\u5230,
\[\sqrt[3]{9abc(a^2+b^2+c^2)}=\sqrt[3]{3\sqrt{abc} \cdot 3\sqrt{abc}(a^2+b^2+c^2)} \]
\[\le \frac{3\sqrt{abc}+3\sqrt{abc}+a^2+b^2+c^2}{3}\le \frac{2(ab+bc+ca)+a^2+b^2+c^2}{3}=3\].
\u4ece\u800c\u8bc1\u660e\u4e86$abc(a^2+b^2+c^2)\le 3$.\u5373\u6240\u9700\u7684\u4e0d\u7b49\u5f0f.
1. rcosx+r^2*cos2x= 2r^2*(cosx)^2+rcosx-r^2=2(rcosx+1/4)^2-r^2-1/8≥-1/4-1/8= -3/8
当且仅当 rcosx=-1/4且|r|=1/2时取等号
2.数学归纳法
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