陈洪葛的博客

设$a,b,c>0$,$a^2+b^2+c^2=3$证明
\[ \sqrt{\frac{a}{a^2+b^2+1}}+\sqrt{\frac{b}{b^2+c^2+1}}+\sqrt{\frac{c}{c^2+a^2+1}}\leq \sqrt{3}\]
证明
由Cauchy-Schwarz不等式
\[ \left(\sum\sqrt{\frac{a}{a^2+b^2+1}}\right)^{2}\leq \sum(a^2+c^2+1)\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}=9\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}\]
下面证明
\[\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}\leq \frac{1}{3} \]
\[ \Leftrightarrow 12\sum{a}-3\sum{a^3}\leq 34-a^2b^2c^2-2\sum{a^4}\]
由$a^2b^2c^2\leq 1$,只要证明
\[ 12\sum{a}-3\sum{a^3}\leq 33-2\sum{a^4}\]
\[ \Leftrightarrow \sum_{cyc}(1-a^2)\left(2a^3-3a+2+\frac{9}{a+1}\right)\geq 0 \]
记$ \sum(1-a^2)S_{a}$,易得$S_{c}\geq S_{b}\geq S_{a} $
由切比雪夫不等式
\[ \sum(1-a^2)S_{a}\geq \frac{1}{3}\sum(1-a^2)\sum S_{a}=0 \]
Done!

http://tieba.baidu.com/p/2558178633?pid=38027005475&cid=&from=prin#38027005475&qq-pf-to=pcqq.temporaryc2c

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3195833#p3195833
 

好久没更新博客了，这段时间比较忙。以后争取有空多贴点好的问题，多谢大家光临我的博客。

Problem

If $a,b,c$ are positive real numbers such that $a+b+c=3$,then
\[ \sqrt{a^3+3b}+\sqrt{b^3+3c}+\sqrt{c^3+3a}\geq 6 \]

Proof. By the Cauchy-Schwarz inequality,we have
\[ (a^3+3b)(a+3b)\geq (a^2+3b)^2 \]
Thus,it suffices to show that
\[ \sum\frac{a^2+3b}{\sqrt{a+3b}}\geq 6 \]
By Holder inequality,we have
\[\left(\sum\frac{a^2+3b}{\sqrt{a+3b}}\right)^{2}\left[\sum(a^2+3b)(a+3b)\right]\geq \left[\sum(a^2+3b)\right]^3=\left(\sum{a^2}+9\right)^3 \]
Therefore,it is enough to show that
\[ \left(\sum a^2+9\right)^3\geq 36\sum(a^2+3b)(a+3b) \]
Let $p=a+b+c=3$ and $q=ab+bc+ca, q\leq 3$.We have
\[ \sum a^2+9=p^2-2q+9=2(9-q) \]
\begin{align*}
\sum(a^2+3b)(a+3b)&=\sum a^3+3\sum a^2b+9\sum a^2+3\sum ab\\
&=(p^3-3pq+3abc)+3\sum a^2b+9(p^2-2q)+3q \\
&=108-24q+3\left(abc+\sum a^2b \right) 
\end{align*}
Since
\[ a^2b+b^2c+c^2a+abc\leq \frac{4}{27}(a+b+c)^3 \]
We get
\[ \sum{(a^2+3b)(a+3b)}\leq 24(5-q) \]
Thus.it suffices to show that
\[ (9-a)^3\geq 108(5-a) \]
Which is equivalent to the obvious inequality
\[ (3-q)^2(21-q)\geq 0 \]
The equality holds for $a=b=c=1 $.     

 Problem

设$a,b,c \in \mathbf{R}$.且$ab+bc+ca=11$.求
\[ \sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91} \]
的最小值。

Solution

由Cauchy-Schwarz

\[ [(a^2+1)+20](5+20)\geq \left[\sqrt{5(a^2+1)}+20\right]^{2}   \]
 \[      [ 2(b^2+1)+12](4+12)\geq \left[\sqrt{8(b^2+1)}+12\right]^2   \]
 \[      [(c^2+1)+90](10+90)\geq \left[\sqrt{10(c^2+1)}+90\right]^2 \]
得到
\[ \sqrt{a^2+21}\geq \sqrt{\frac{1}{5}\left(a^2+1\right)}+4 \]
\[ \sqrt{2b^2+14}\geq \sqrt{\frac{1}{2}\left(b^2+1\right)}+3 \]
\[ \sqrt{c^2+91} \geq \sqrt{\frac{1}{10}\left(c^2+1\right)}+9 \]
\begin{align*}
\sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}&\geq 16 +\sqrt{\frac{1}{5}\left(a^2+1\right)}+\sqrt{\frac{1}{2}\left(b^2+1\right)}+\sqrt{\frac{1}{10}\left(c^2+1\right)}\\
&\geq 16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}
\end{align*}
注意到恒等式
\[ (a^2+1)(b^2+1)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2\geq (ab+bc+ca-1)^2\ge 100 \]
故
\[ 16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}\geq 19 \]
\[ \sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}\geq 19 \]
等号成立当$a=2,b=1,c=3 $                                                        $\square$
 

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 $a=b=c=1$.    $\square$

__________________________________________________________________________________________________

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!