证明：当 x > 0 x>0 x>0时， sin x + cos x > − x 2 + x + 1 \sin x+\cos x>-x^2+x+1 sinx+cosx>−x2+x+1
证明：
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令
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f(x)=\sin x+\cos x+x^2-x-1
f(x)=sinx+cosx+x2−x−1
\qquad
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f'(x)=\cos x-\sin x+2x-1
f′(x)=cosx−sinx+2x−1
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\qquad f''(x)=-\sin x-\cos x+2=(1-\sin x)+(1-\cos x)
f′′(x)=−sinx−cosx+2=(1−sinx)+(1−cosx)
\qquad 当 x > 0 x>0 x>0时， f ′ ′ ( x ) ≥ 0 f''(x)\geq0 f′′(x)≥0，即 f ′ ( x ) f'(x) f′(x)单调递增，故 f ′ ( x ) f'(x) f′(x)有最小值 f ′ ( 0 ) = 0 f'(0)=0 f′(0)=0
\qquad 故 f ′ ( x ) f'(x) f′(x)有最小值 f ′ ( 0 ) = 0 f'(0)=0 f′(0)=0，即恒有 f ( x ) > f ( 0 ) = 0 f(x)>f(0)=0 f(x)>f(0)=0
f ( x ) \qquad f(x) f(x)单调递增，即 f ( x ) f(x) f(x)有最小值 f ( 0 ) = 0 f(0)=0 f(0)=0
x > 0 \qquad x>0 x>0时 f ( x ) > 0 f(x)>0 f(x)>0，得证 x > 0 x>0 x>0时 sin x + cos x > − x 2 + x + 1 \sin x+\cos x>-x^2+x+1 sinx+cosx>−x2+x+1