Mar 7

      设$f(x)$是定义在R上的实函数,且对$\forall x$有$f(x),f'(x),f''(x),f'''(x)>0$,假设对$\forall x$有$f'''(x)<f(x)$证明:对一切$x$都有$f'(x)<2f(x)$.
 

证明 对任意固定的$c$,令
\[g(x)=f(x)-f'(x)(x-c)+\frac{f''(x)}{2}(x-c)^{2}\]

\[ g'(x)=\frac{f'''(x)}{2}(x-c)^{2}\geq 0 \]
所以$g(x)$是递增函数,则对$\forall y>0$
\[ g(c+y)>g(c-y) \]
也就是
\[ g(c+y)=f(c+y)-yf'(c+y)+\frac{f''(c+y)}{2}y^{2}>f(c-y)+yf'(c-y)+\frac{f''(c-y)}{2}y^{2}=g(c-y)\]
注意到
\[ f(c+y)-yf'(c+y)+\frac{f''(c+y)}{2}y^{2}>f(c-y)+yf'(c-y)+\frac{f''(c-y)}{2}y^{2}>\frac{f''(c-y)}{2}y^{2} \]
故有
\[ f(c+y)-yf'(c+y)>\frac{1}{2}y^{2}[f''(c-y)-f''(c+y)] \]
由拉格朗日中值定理知存在$\theta\in(c-y,c+y)$,满足
\[ f''(c+y)-f''(c-y)=2yf'''(\theta)\leq 2yf(\theta)<2yf(c+y) \]
所以
\[ f(c+y)-yf'(c+y)+y^3f(c+y)>0 \]
而由条件可推断
\[\frac{1+y^3}{y}f(c+y)>f'(c+y) \]
当$y=\frac{1}{\sqrt[3]{2}}$上面的$ \frac{1+y^3}{y}=\frac{3}{\sqrt[3]{4}}<2,$
故命题得证。

下面一种证明来着风碎便士同学。
令$c\geq 0$是$f(x)$的下确界,则$f(x)$在$x\rightarrow -\infty$时单调递于$c$,并且容易得到$f'(x),f''(x)$在$x\rightarrow -\infty$时均单调趋于$0$.
这是由于$f''(x)>0$说明$f'(x)$是递增的,故$\displaystyle \lim_{x\rightarrow -\infty}{f'(x)}$要么为$-\infty$,要么为一个常数,而由$\displaystyle \lim_{x\rightarrow -\infty}{f(x)}=c$知,$\displaystyle \lim_{x\rightarrow -\infty}{f'(x)}=0$(其他情况均会矛盾),同样可说明$\displaystyle \lim_{x\rightarrow -\infty}{f''(x)}=0$
从而
\[ f'''(x)f''(x)<f(x)f''(x)<f(x)f''(x)+(f'(x))^2 \]
两边从$a$到$x$积分,得
\[ \frac{1}{2}(f''(x))^2-\frac{1}{2}(f''(a))^2<f(x)f'(x)-f(a)f'(a) \]
再令$a\rightarrow -\infty$得到
\[ (f''(x))^2\leq 2f(x)f'(x) \]
从而
\[ f'''(x)(f''(x))^2\leq 2f^{2}(x)f'(x) \]
两边从$a$到$x$积分,再令$a\rightarrow -\infty$得到
\[ (f''(x))^3\leq 2f^{3}(x) \]
\[ f''(x)\leq \sqrt[3]{2}f(x) \]
\[ f'(x)f''(x) \leq \sqrt[3]{2}f(x)f'(x) \]
两边从$a$到$x$积分,再令$a\rightarrow -\infty$得到
\[ f'(x)\leq \sqrt[6]{2}f(x)<2f(x) \]

Mar 3

设$f(x)$是正的递减的函数,证明

\[ \int_{0}^{1}{xf(x)^2dx} \int_{0}^{1}{f(x)dx}\le \int_{0}^{1}{f(x)^2dx }\int_{0}^{1}{xf(x)dx} \]

证明 考虑

\[ G(x,y)= \frac{1}{2}(x-y)(f(x)-f(y))f(x)f(y)\]

显然有

\[ G(x,y)\leq 0 \]

\[ \iint_{[0,1]^{2}}{G(x,y)dxdy}=\int_{0}^{1}{xf^{2}(x)dx}\cdot \int_{0}^{1}{f(x)dx}-\int_{0}^{1}{xf(x)dx}\cdot\int_{0}^{1}{f^{2}(x)dx}\]

打开即得证

Mar 3

 

计算

\[ \int_{0}^{\infty}{\frac{1}{\sqrt{x(1+e^{x})}}dx}\]

Solution

\begin{align}
\int_0^\infty\frac{1}{\sqrt{x(1+e^x)}}\mathrm{d}x
&=2\int_0^\infty\frac{1}{\sqrt{1+e^{x^2}}}\mathrm{d}x\\
&=2\int_0^\infty(1+e^{-x^2})^{-1/2}e^{-x^2/2}\;\mathrm{d}x\\
&=2\int_0^\infty\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}e^{(2k+1)x^2/2}\;\mathrm{d}x\\
&=\sum_{k=0}^\infty(-\tfrac{1}{4})^k\binom{2k}{k}\sqrt{\frac{2\pi}{2k+1}}
\end{align}

 

 

 

Mathematics StackExchange

Mar 3

据说还有

Mar 3

早上起来看见kuing粉丝群里的天书同学发了一个有意思的不等式,经过一番思考后弄了出来。
Problem:
Let $a,b,c$ be postive numbers,with $abc=1$,show that
\[ \frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}\leq \frac{3}{4} \]
证明
先用$ a=\frac{y}{x},b=\frac{z}{y},c=\frac{x}{z} $替换,得到
\[ \frac{xy}{3x^2+y^2}+\frac{yz}{3y^2+z^2}+\frac{xz}{3z^2+x^2}\leq \frac{3}{4} \]

\[ \frac{2x^2+(x-y)^2}{3x^2+y^2}+\frac{2y^2+(y-z)^2}{3y^2+z^2}+\frac{2z^2+(z-x)^2}{3z^2+x^2}\geq \frac{3}{2} \]
于是,自然的运用Cauchy-Schwarz,得到
\[ \sum{\frac{2x^2+(x-y)^2}{3x^2+y^2}}\geq \frac{\left(\sum{\sqrt{2x^2+(x-y)^2 }}\right)^{2}}{4\sum{x^2}}\]
故只要证明
\[ \frac{\left(\sum{\sqrt{2x^2+(x-y)^2 }}\right)^{2}}{4\sum{x^2}}\geq \frac{3}{2}\]
经过一番化简,得到
\[ \sum{\sqrt{[2x^2+(x-y)^2][2y^2+(z-y)^2]}}\geq \sum{x^2}+\sum{xy} \]
显然,再次运用Cauchy-Schwarz马上得到答案
\[ \sqrt{[2x^2+(x-y)^2][2y^2+(z-y)^2]}\geq 2xy+(y-x)(y-z) \]
只要把上面类似的3个式子相加就好。
由此,不等式得证。