陈洪葛的博客

Problem 1:Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3.$ Prove that
\[   \sqrt{a^3+2bc}+\sqrt{b^3+2ca}+\sqrt{c^3+2ab} \geq 3\sqrt{3}.\]

Proof：by Holder inequality,we have
\[\left(\sum{\sqrt{a^3+2bc}}\right)^2\cdot\left[\sum\frac{(a^2+2bc)^3}{a^3+2bc}\right]\geq (a+b+c)^6 \]
Therefore,it's suffice to prove
\[\sum\frac{(a^2+2bc)^3}{a^3+2bc}\leq 27 \]
by Holder inequality,we can see that
\[ (a+2bc)(a^3+2bc)(a^2+2bc)\geq (a^2+2bc)^3 \]
which imply
\[ \frac{(a^2+2bc)^3}{a^3+2bc}\leq (a+2bc)(a^2+2bc) \]
Thus,it's enough to prove
\[ \sum{(a+2bc)(a^2+2bc)}\leq 27 \]
After homogenous,it's
\[ (\sum{a})(\sum{a^3}+6abc)+12\sum{a^2b^2}+6\sum{a^2bc}\leq (a+b+c)^4 \]
Or
\[ ab(a-b)^2+bc(b-c)^2+ca(c-a)^2\geq 0 \]
Which is obviously true,Hence we are done! Equality occurs when $a=b=c=1$.    $\square$

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Problem 2:      If $a,b,c$ are positive real numbers such that $a+b+c=3,$ then
\[ \dfrac{a}{12a+5b^2}+\dfrac{b}{12b+5c^2}+\dfrac{c}{12c+5a^2} \leq \dfrac{3}{17}.\]

Proof:First,we can rewrite the inequality into
\[ \frac{b^2}{4a^2+5b^2+4ab+4ac}+\frac{c^2}{4b^2+5c^2+4bc+4ba}+\frac{a^2}{4c^2+5a^2+4ca+4bc}\geq \frac{3}{17}\]
by Cauchy-Schwarz inequality,we have
\[\left(\sum_{cyc}{\frac{b^2}{4a^2+5b^2+4ab+4ac}}\right)\left[\sum_{cyc}{(2b+c)^2(4a^2+5b^2+4ab+4ac)}\right]\geq \left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\]
Therefore,it's suffice to prove that
\[ 17\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\geq 3\left[\sum_{cyc}{(2b+c)^2(4a^2+5b^2+4ab+4ac)}\right]\]
after some computation,it's
\[ 8\sum_{cyc}{a^4}+20\sum_{cyc}{ab^3}-4\sum_{cyc}{a^3b}-102\sum{a^2bc}+78\sum{a^2b^2}\geq 0 \]
Now,Using Vo Quoc Ba Can's sum of square technique,for $m=8,n=78,p=20,g=-4$,we have to check
\[ \left\{\begin{array}{ll}
        3m(m+n)\geq p^2+pg+g^2 ,\\
        m>0 ,
     \end{array}
\right.\]
it can be easily see that $ 3m(m+n)=2064>336=p^2+pg+g^2 $,Hence the inequality is holds.
Done!          $\square$
 

For $a,b,c$ are real numbers such that $ab+bc+ca>0$ with $a+b+c=1 $,find min
\[ P=\frac{2}{|a-b|}+\frac{2}{|b-c|}+\frac{2}{|c-a|}+\frac{5}{\sqrt{ab+bc+ca}} \]

（choisiwon，Japan）

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=539142
Solution (thanks tian275461's hint)
We will prove that
\[ P\geq 10\sqrt{6} \]
Without loss of generally,we can assume that $a>b>c$
by AM-GM inequality
\[ \frac{2}{a-b}+\frac{2}{b-c}\geq \frac{8}{a-c} \]
which leads
\[ P\geq \frac{10}{a-c}+\frac{5}{\sqrt{ab+bc+ca}}\geq \frac{4}{\frac{a-c}{10}+\frac{\sqrt{ab+bc+ca}}{5}}\]
Thus,it's suffice to prove
\[ \frac{a-c}{10}+\frac{\sqrt{ab+bc+ca}}{5}\leq \frac{4}{10\sqrt{6}}=\frac{1}{15}\sqrt{6}\]
Or
\[ (a-c)+2\sqrt{(a+c)-(a+c)^2+ac}\leq \frac{2}{3}\sqrt{6} \]
Now,Using Cauchy-Schwarz and AM-GM inequality,we have
\begin{align*}
 (a-c)+2\sqrt{(a+c)-(a+c)^2+ac}&\leq \sqrt{2[(a-c)^2+4(a+c)-4(a+c)^2+4ac]}\\
&=\sqrt{2(a+c)[4-3(a+c)]}\\
&=\frac{\sqrt{2}}{\sqrt{3}}\sqrt{3(a+c)[4-3(a+c)]}\\
&\leq \frac{2}{3}\sqrt{6}
\end{align*}
Done!

Let $a,b,c\geq 0$ with $ a+b+c=3$ ,show that

\[ \sqrt{a^2+bc+2}+\sqrt{b^2+ca+2}+\sqrt{c^2+ab+2}\geq 6 \]

(lhclp008)

Proof

by Holder inequality
\[ \left(\sum{\sqrt{a^2+bc+2}}\right)^2\left[\sum{\frac{(a^2+2bc+9)^3}{a^2+bc+2}}\right]\geq [(a+b+c)^2+27]^3=36^3 \]
Therefore,it's suffice to check
\[  \frac{(a^2+2bc+9)^3}{a^2+bc+2}+\frac{(b^2+2ca+9)^3}{b^2+ca+2}+\frac{(c^2+2ab+9)^3}{c^2+ab+2}\leq 1296 \]

Which can be checked by Muirhead's theorem.

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also,we have

\[ \sqrt{3a^2+4bc+9}+\sqrt{3b^2+4ca+9}+\sqrt{3c^2+4ab+9}\geq 12. \]

in the same conditions.

(quyhktn-qa1)

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=538752

it's suffices to prove

$$ \frac{(a^2+2bc+9)^3}{3a^2+4bc+9}+\frac{(b^2+2ca+9)^3}{3b^2+4ca+9}+\frac{(c^2+2ab+9)^3}{3c^2+4ab+9}\le 324 $$

Which can also be checked by Muirhead's theorem.

证明
\[ \lim_{n\to\infty}{\int_{0}^{\frac{\pi}{2}}{\sin{t^{n}}dt}}=0 \]

证
\[ I=\int_{0}^{\frac{\pi}{2}}{\sin{t^{n}}dt}=\frac{1}{n}\cdot\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{y^{\frac{1}{n}-1}\sin{y}dy}\qquad (y=t^{n}) \]
而
\[ \Gamma\left(1-\frac{1}{n}\right)=\int_{0}^{\infty}{u^{-\frac{1}{n}}e^{-u}du}=\int_{0}^{\infty}{y^{1-\frac{1}{n}}\cdot x^{-\frac{1}{n}}e^{-xy}dx} \]
\begin{align*}
 I&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{\left(\int_{0}^{+\infty}{x^{-\frac{1}{n}}e^{-xy}dx}  \right)\cdot\sin{y}dy}\\
&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{+\infty}{x^{-\frac{1}{n}}\cdot\left(\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{e^{-xy}\sin{y}dy}  \right)dx}\\
&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{+\infty}{\frac{x^{-\frac{1}{n}}}{1+x^2}dx}-\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{\infty}{\frac{x^{-\frac{1}{n}}\left[\cos\left(\frac{\pi}{2}\right)^{n}+\left(\frac{\pi}{2}\right)^{n}\sin\left(\frac{\pi}{2}\right)^{n}  \right] }{(1+x^2)e^{x\left(\frac{\pi}{2}\right)^{n}}}dx}\\
&=I_{1}-I_{2}
\end{align*}
而我们知道
\[ \int_{0}^{+\infty}{\frac{t^{a-1}}{1+t}dt}=\frac{\pi}{\sin{\pi a}}\qquad  (\text{Euler}) \]
\[ \Rightarrow \int_{0}^{+\infty}{\frac{x^{-\frac{1}{n}}}{1+x^2}dx}=\frac{\pi}{2\cos{\frac{1}{2n}}}\]
\[ \Rightarrow \lim_{n\to\infty}{I_{1}}=0 \]
\[ I_{2}=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot\left(\int_{0}^{1}+\int_{1}^{+\infty}\right)=S_{1}+S_{2}\]
\begin{align*}
|S_{1}|&=\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\left|\int_{0}^{1}{\frac{x^{-\frac{1}{n}}\left[\cos\left(\frac{\pi}{2}\right)^{n}+\left(\frac{\pi}{2}\right)^{n}\sin\left(\frac{\pi}{2}\right)^{n}  \right] }{(1+x^2)e^{x\left(\frac{\pi}{2}\right)^{n}}}dx} \right|\\
&\leq\frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\int_{0}^{1}{\frac{[\left(\frac{\pi}{2}\right)^{n}+1]x^{-\frac{1}{n}}}{e^{x\left(\frac{\pi}{2}\right)^{n}}}dx}\\
&= \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot \frac{\left(\frac{\pi}{2}\right)^{n}+1}{\left(\frac{\pi}{2}\right)^{n-1}}\int_{0}^{\left(\frac{\pi}{2}\right)^{n}}{z^{-\frac{1}{n}}e^{-z}dz}\\
&\leq \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot \frac{\left(\frac{\pi}{2}\right)^{n}+1}{\left(\frac{\pi}{2}\right)^{n-1}}\int_{0}^{+\infty}{z^{-\frac{1}{n}}e^{-z}dz}\\
&=\frac{1}{n}\cdot\left[\left( \frac{\pi}{2}\right)+\left( \frac{\pi}{2}\right)^{1-n}  \right]\to 0
\end{align*}
\begin{align*}
 |S_{2}|&= \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\left|\int_{1}^{\infty}{\frac{x^{-\frac{1}{n}}\left[\cos\left(\frac{\pi}{2}\right)^{n}+\left(\frac{\pi}{2}\right)^{n}\sin\left(\frac{\pi}{2}\right)^{n}  \right] }{(1+x^2)e^{x\left(\frac{\pi}{2}\right)^{n}}}dx}\right|\\
&\leq \frac{1}{n\Gamma\left(1-\frac{1}{n}\right)}\cdot \int_{1}^{+\infty}{\frac{x^{-\frac{1}{n}}}{1+x^2}dx}\\
&\leq \frac{\pi}{2n\Gamma\left(1-\frac{1}{n}\right)}\to 0
\end{align*}
故
\[ \lim_{n\to\infty}{\int_{0}^{\frac{\pi}{2}}{\sin{t^{n}}dt}}=0 \]

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顺便计算
\begin{align*}
\int_0^\infty \sin \left( x^n\right)dx &= \frac{1}{n}\int_0^\infty x^{\frac{1}{n}-1} \sin(x) \ dx \quad (x^n \mapsto x) \\
&= \frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)}\int_0^\infty \left(\int_0^\infty u^{-\frac{1}{n}}e^{-xu}du\right) \sin(x) \ dx \\
&= \frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)} \int_0^\infty u^{-\frac{1}{n}} \left( \int_0^\infty e^{-xu}\sin(x) \ dx\right)du \\
&= \frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)} \int_0^\infty \frac{u^{-\frac{1}{n}}}{1+u^2}du \\
&= \frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)} \int_0^{\frac{\pi}{2}}\tan^{-\frac{1}{n}}(\theta) d\theta \quad (u=\tan \theta) \\
&= \frac{1}{n \Gamma \left( 1-\frac{1}{n}\right)}\int_0^{\frac{\pi}{2}}\sin^{-\frac{1}{n}}(\theta) \cos^{\frac{1}{n}}(\theta) d\theta \\
&= \frac{1}{2n \Gamma \left( 1-\frac{1}{n}\right)} \mathrm{B} \left( \frac{1-n}{2},\frac{1+n}{2}\right) \\
&= \frac{1}{2n \Gamma \left( 1-\frac{1}{n}\right)} \Gamma \left( \frac{n-1}{2n}\right)\Gamma \left( \frac{n+1} {2n}\right) \\
&= \frac{\sin \left( \frac{\pi}{n}\right)}{2n\cos \left( \frac{\pi}{2n}\right)}\Gamma \left( \frac{1}{n}\right) \\
&= \frac{1}{n}\sin \left(\frac{\pi }{2n} \right)\Gamma \left( \frac{1}{n}\right)
\end{align*}

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显然，上面的证明弄复杂了，太非主流了。下面给出个主流的证明

对$\forall \varepsilon>0$,存在$ 0<a<\frac{\varepsilon}{4} $,注意到
\begin{align*}
\int_{0}^{\frac{\pi}{2}}{\sin{t^n}dt}&= \int_{0}^{1-a}{\sin{t^n}dt}+ \int_{1-a}^{1+a}{\sin{t^n}dt}+ \int_{1+a}^{\frac{\pi}{2}}{\sin{t^n}dt}\\
&=I_{1}+I_{2}+I_{3}
\end{align*}
由$ |\sin{x}|\leq 1 $知
\[ |I_{2}|<\frac{\varepsilon}{2} \]
对$I_{1}$,有
\[ |I_{1}|\leq \int_{0}^{1-a}{|\sin{t^n}|dt}\leq \int_{0}^{1-a}{t^{n}dt}\leq \frac{(1-a)^{n+1}}{n+1} \]
显然可以找到一个$N_{1}>0$使得$n>N_{1}$时有
\[ |I_{1}|\leq \frac{\varepsilon}{4} \]
而对$I_{3}$
\begin{align*}
 I_{3}&=\int_{1+a}^{\frac{\pi}{2}}{\sin{t^n}dt}\\
&= \int_{1+a}^{\frac{\pi}{2}}{\frac{d(-\cos{t^n})}{nt^{n-1}}}\\
&=\frac{-\cos t^n}{nt^{n-1}}\bigg|_{1+a}^{\frac{\pi}{2}}+\frac{1-n}{n}\cdot \int_{1+a}^{\frac{\pi}{2}}{\frac{\cos{t^{n}}}{t^{n}}dt}\\
&=\frac{\cos{(1+a)^n}-\cos{\left(\frac{\pi}{2}\right)^n}}{n(1+a)^{n-1}}+\frac{1-n}{n}\cdot \int_{1+a}^{\frac{\pi}{2}}{\frac{\cos{t^{n}}}{t^{n}}dt}
\end{align*}
显然存在$N_{2}$使得当$n>N_{2}$时有
\[ |I_{3}|<\frac{\varepsilon}{4} \]
这样，取$N=\max\{N_{1},N_{2}\} $,当$n>N$时，就有
\[ |I|<\varepsilon \]
\[ \Rightarrow \lim_{n\to\infty}{\int_{0}^{\frac{\pi}{2}}{\sin{t^n}}dt}=0 \]

设$x,y,z$是三个不同的实数，证明
\[ \frac{1}{\max\{x,y,z\}}\leq \frac{2x\ln{x}}{(x-y)(x-z)}+\frac{2y\ln{y}}{(y-x)(y-z)}+\frac{2z\ln{z}}{(z-y)(z-x)}\leq  \frac{1}{\max\{x,y,z\}} \]
（西西）
证明：（lhclp008）
不失一般性，我们假设$x\geq y\geq z$,注意到对任意的$a,b>0$有不等式
\[ \frac{1}{\sqrt{ab}}\geq \frac{\ln{a}-\ln{b}}{a-b}\geq \frac{2}{a+b} \]
于是
\begin{align*}
\frac{1}{x}&\leq \frac{1}{x}+\frac{(\frac{2xy}{x+y}-z)(x-y)}{xy(x-z)}\\
&=\frac{\frac{4x}{x+y}-\frac{(y+z)}{y}}{(x-z)}\\
&\leq \frac{\frac{4x}{x+y}-2\frac{\sqrt{yz}}{y}}{x-z}\\
&=\frac{2x(y-z)\cdot\frac{2(x-y)}{x+y}+2z(y-x)\frac{(y-z)}{\sqrt{yz}}}{(x-y)(x-z)(y-z)}\\
&\leq \frac{2x(y-z)(\ln{x}-\ln{y})+2z(y-x)(\ln{y}-\ln{z})}{(x-y)(x-z)(y-z)}\\
&=\frac{2x\ln{x}}{(x-y)(x-z)}+\frac{2y\ln{y}}{(y-x)(y-z)}+\frac{2z\ln{z}}{(z-y)(z-x)}
\end{align*}
\begin{align*}
&\frac{2x\ln{x}}{(x-y)(x-z)}+\frac{2y\ln{y}}{(y-x)(y-z)}+\frac{2z\ln{z}}{(z-y)(z-x)}\\
&\leq \frac{2x(y-z)\frac{x-y}{\sqrt{xy}}+2z(y-x)\frac{2(y-z)}{y+z}}{(x-y)(x-z)(y-z)}\\
&=-\frac{-2z\frac{\sqrt{xy}}{y}+\frac{4z^2}{y+z}+x-z}{z(x-z)}+\frac{1}{z}\\
&\leq -\frac{-z\cdot\frac{x+y}{y}+\frac{4z^2}{y+z}+x-z}{z(x-z)}+\frac{1}{z}\\
&=-\frac{xy+xz-2yz}{yz(y+z)(x-z)}\cdot (y-z)+\frac{1}{z}\\
&\leq \frac{1}{z} 
\end{align*}

Reference:http://tieba.baidu.com/p/2388204458