设 $f$ 为 $[0,1]$ 上的连续正函数, 且 $dps{f^2(t)leq 1+2int_0^t f(s) d s}$. 证明: $f(t)leq 1+t$.
证明: 设 $dps{F(t)=int_0^t f(s) d s}$, 则 $F(0)=0$, 且 $$eex ea F'^2(t)&leq 1+2F(t),\ cfrac{ d F(t)}{sqrt{1+2F(t)}}&leq d t,\ sqrt{1+2F(t)}-sqrt{1+2F(0)}&leq t,\ sqrt{1+2F(t)}&leq 1+t,\ 2F(t)&leq (1+t)^2-1=2t+t^2,\ f^2(t)&leq 1+2F(t)=1+2t+t^2=(1+t)^2,\ f(t)&leq 1+t. eea eeex$$