一道积分不等式
2018.05.11
\(f\)是正值函数，\(f\in C[a,b]\),记\(\int_a^b{f\left( x \right)}dx=A\),证明：\(\int_a^b{f\left( x \right) e^{f\left( x \right)}}dx\int_a^b{\frac{1}{f\left( x \right)}}dx\ge \left( b-a \right) \left( b-a+A \right) .\)
\(solution:\)
由\(Schwarz不等式\)，左边\(\ge\left( \int_a^b{\sqrt[]{e^{f\left( x \right)}}}dx \right) ^2,\)
根据\(Taylor\)公式，\(e^{f\left( x \right)}=1+f\left( x \right) +\frac{1}{2}f^2\left( x \right) +\cdots \ge \left( 1+\frac{1}{2}f\left( x \right) \right) ^2,\)
故左边\(\left( \int_a^b{\left( 1+\frac{1}{2}f\left( x \right) \right) dx} \right) ^2=\left( b-a+\frac{1}{2}A \right) ^2\ge \left( b-a \right) \left( b-a+A \right) .\)
本题解答由kkd给出.