关于不等式的证明题

1.c^3=(a^2+b^2)^1.5
=>(c^3)^2=c^6=(a^2+b^2)^3=a^6+b^6+3a^4b^2+3b^4a^2
(a^3+b^3)^2=a^6+b^6+2a^3b^3
=>(c^3)^2-(a^3+b^3)^2=3a^4b^2+3b^4a^2-2a^3b^3
=a^3b^3(3(a/b+b/a)-2)
>=a^3b^3(3*2-2)>0
=>(c^3)^2>(a^3+b^3)^2
=>c^3>a^3+b^3

欲证：(c^3)/2<a^3+b^3
须证：(a^2+b^2)^1.5<2(a^3+b^3)
须证：(a^2+b^2)^3<4(a^3+b^3)^2
须证：a^6+b^6+3a^4b^2+3a^2b^4<4a^6+4b^6+8a^3b^3
须证：3a^4b^2+3a^2b^4<3a^6+3b^6+8a^3b^3
须证：3(a/b+b/a)<3(a^3/b^3+b^3/a^3)+8
须证：3(a/b+b/a)<3(a/b+b/a)(a^2/b^2+b^2/a^2-1)+8
须证：0<3(a/b+b/a)(a^2/b^2+b^2/a^2-2)+8
须证：0<3(a/b+b/a)(a/b-b/a)^2+8
显然成立
因此原题成立

2.1/n^2<1/(n^2-1)
=>1/9<1/8=1/2*(1/2-1/4)
=>1/25<1/24=1/2*(1/4-1/6)
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.
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=>1/[(2n+1)^2]<1/2n(2n+2)=1/2*(1/2n-1/(2n+2))
=>1/9+1/25+.....<1/2*(1/2-1/4+1/4-1/6+...1/2n-1/(2n+2))=1/2 * (1/2-1/(2n+2))&l