设f(x)在[0,1]上连续且可导,又f(0)=0,0≤f'(x)≤1 试证:[ ∫^(0,1)f(x)dx]^2≥∫^(0,1)[f(x)]^3dx虽然想从0≤f'(x)≤1入手说明f(x)>[f(x)]^3,但貌似没什么用

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设f(x)在[0,1]上连续且可导,又f(0)=0,0≤f'(x)≤1 试证:[ ∫^(0,1)f(x)dx]^2≥∫^(0,1)[f(x)]^3dx虽然想从0≤f'(x)≤1入手说明f(x)>[f(x)]^3,但貌似没什么用

设f(x)在[0,1]上连续且可导,又f(0)=0,0≤f'(x)≤1 试证:[ ∫^(0,1)f(x)dx]^2≥∫^(0,1)[f(x)]^3dx虽然想从0≤f'(x)≤1入手说明f(x)>[f(x)]^3,但貌似没什么用
设f(x)在[0,1]上连续且可导,又f(0)=0,0≤f'(x)≤1 试证:[ ∫^(0,1)f(x)dx]^2≥∫^(0,1)[f(x)]^3dx
虽然想从0≤f'(x)≤1入手说明f(x)>[f(x)]^3,但貌似没什么用

设f(x)在[0,1]上连续且可导,又f(0)=0,0≤f'(x)≤1 试证:[ ∫^(0,1)f(x)dx]^2≥∫^(0,1)[f(x)]^3dx虽然想从0≤f'(x)≤1入手说明f(x)>[f(x)]^3,但貌似没什么用
两边同时求导
或相减后求导

只证明出f(x)<=1后面怎么证明就不会了