只需證明1+1/根號(hào)2^3+1/根號(hào)3^3+.+1/根號(hào)k^3≤3-2/根號(hào)k
用數(shù)學(xué)歸納法:k=1時(shí),左=1,右=1,成立,
設(shè)k=n時(shí)原不等式成立,則k=n+1時(shí),左=1+1/根號(hào)2^3+1/根號(hào)3^3+.+1/根號(hào)n^3+1/根號(hào)(n+1)^3
≤3-2/根號(hào)n+1/根號(hào)(n+1)^3
下證-2/根號(hào)n+1/根號(hào)(n+1)^3≤-2/根號(hào)(n+1)
而2/根號(hào)n-2/根號(hào)(n+1)=2[根號(hào)(n+1)-根號(hào)n]/根號(hào)n(n+1)=2/[根號(hào)(n+1)+根號(hào)n]根號(hào)n(n+1)
≤2/2根號(hào)(n+1)^3=1/根號(hào)(n+1)^3
所以n=k+1時(shí),左=1+1/根號(hào)2^3+1/根號(hào)3^3+.+1/根號(hào)n^3+1/根號(hào)(n+1)^3
≤3-2/根號(hào)n+1/根號(hào)(n+1)^3≤3-2/根號(hào)(n+1),所以k=n+1時(shí)成立
綜上,有1+1/根號(hào)2^3+1/根號(hào)3^3+.+1/根號(hào)k^3≤3-2/根號(hào)k
而3-2/根號(hào)k<3恒成立(k∈Z),所以
1+1/根號(hào)2^3+1/根號(hào)3^3+.+1/根號(hào)n^3<32/[根號(hào)(n+1)+根號(hào)n]根號(hào)n(n+1)≤2/2根號(hào)(n+1)^3??額,不好意思,是≥根號(hào)(n+1)+根號(hào)n≤2根號(hào)(n+1)根號(hào)n≤根號(hào)(n+1)根號(hào)(n+1)≤根號(hào)(n+1)3個(gè)式子相乘有[根號(hào)(n+1)+根號(hào)n]根號(hào)n(n+1)≤2根號(hào)(n+1)^3所以2/[根號(hào)(n+1)+根號(hào)n]根號(hào)n(n+1)≥2/2根號(hào)(n+1)^3