设a、b、c为正数,且a^2+b^2+c^2=3,证明:1/(1+2ab)+1/(1+2bc)+1/(1+2ca)>=1.a、b、c为正数,故1/(1+2ab)+1/(1+2bc)+1/(1+2ca)≥1/(1+a^2+b^2)+1/(1+b^2+c^2)+1/(1+c^2+a^2)≥3*{3/[(a^2+b^2)+(b^2+c^2)+(c^2+a^2)+3]}=9/[3+2(a^2+b^2+c^2)]=1即1/

Question

设a、b、c为正数,且a^2+b^2+c^2=3,证明:1/(1+2ab)+1/(1+2bc)+1/(1+2ca)>=1.
a、b、c为正数,故
1/(1+2ab)+1/(1+2bc)+1/(1+2ca)
≥1/(1+a^2+b^2)+1/(1+b^2+c^2)+1/(1+c^2+a^2)
≥3*{3/[(a^2+b^2)+(b^2+c^2)+(c^2+a^2)+3]}
=9/[3+2(a^2+b^2+c^2)]
=1
即1/(1+2ab)+1/(1+2bc)+1/(1+2ca)≥1.
如果你会，帮我细细讲讲好么，

Answer 1

主要用到两个不等式
a²+b²≥2ab
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真分数相加后结果≥分子分母分别相加后结果*真分数个数
即a/b+m/n≥2(a+m)/(b+n)其中a,b,m,n为正数,≥1,a