陈洪葛的博客

Let$ a,b,c>0 $with $a^3+b^3+c^3=a^4+b^4+c^4$,Prove that
\[\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+c^3+a^3}+\frac{c}{c^2+a^3+b^3}\geq 1 \]
(2012 Turkey)
Proof:
by Cauchy-Schwarz,we have
\[\left(\sum{\frac{a}{a^2+b^3+c^3}}\right)\left[\sum{a(a^2+b^3+c^3)}\right]\geq (a+b+c)^2\]
Therefore,it's suffice to prove
\[(a+b+c)^2\geq \sum{a(a^2+b^3+c^3)}\]
Or
\[(a+b+c)^2\geq (a+b+c)(a^3+b^3+c^3)\]
\[\Leftrightarrow  (a+b+c)(a^4+b^4+c^4)^2\geq (a^3+b^3+c^3)^3\]

Done!

设$a,b,c>0$且$a+b+c=3$,$k>0$,证明
\[ (b+c)\sqrt[k]{\frac{bc+1}{a^2+1}}+(c+a)\sqrt[k]{\frac{ca+1}{b^2+1}}+(a+b)\sqrt[k]{\frac{ab+1}{c^2+1}}\geq 6 \]
证明
不等式等价于
\[ \sum{a\left(\sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\right)}\geq 6 \]
由AM-GM,
\[ k\cdot\sqrt[k]{\frac{ca+1}{b^2+1}}+k\cdot\sqrt[k]{\frac{ab+1}{c^2+1}}+\frac{b^2+1}{ab+1}+\frac{c^2+1}{ca+1}\geq 2(k+1) \]
得到
\[ \sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\geq \frac{2(k+1)}{k}-\frac{1}{k}\left(\frac{b^2+1}{ab+1}+\frac{c^2+1}{ac+1}\right)\]
\[a\left(\sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\right)\geq \frac{2(k+1)a}{k}-\frac{1}{k}\left(\frac{a(b^2+1)}{ab+1}+\frac{a(c^2+1)}{ac+1}\right)\]
\begin{align*}
 \sum{a\left(\sqrt[k]{\frac{ca+1}{b^2+1}}+\sqrt[k]{\frac{ab+1}{c^2+1}}\right)}&\geq  \frac{2(k+1)a}{k}-\frac{1}{k}\left(\frac{a(b^2+1)}{ab+1}+\frac{a(c^2+1)}{ac+1}\right)\\
 &=\frac{6(k+1)}{k}-\frac{1}{k}\left[\sum{\frac{a(b^2+1)}{ab+1}}+\sum{\frac{a(c^2+1)}{ac+1}}\right]\\
 &=6
\end{align*}

设$a,b,c>0$ 证明
\[ \frac{1}{(3a+2b+c)^2}+\frac{1}{(3b+2c+a)^2}+\frac{1}{(3c+2a+b)^2}\leq \frac{1}{4(ab+bc+ca)} \]
证明:
1、$ c\geq b\geq a$
由 AM-GM
\[ \frac{1}{(3a+2b+c)^2}=\frac{1}{[(a+2b)+(c+2a)]^2}\leq \frac{1}{4(a+2b)(c+2a)} \]
\[ \Leftrightarrow \frac{1}{(a+2b)(c+2a)}+\frac{1}{(b+2c)(a+2b)}+\frac{1}{(c+2a)(b+2c)}\leq \frac{1}{ab+bc+ca} \]
\[ \Leftrightarrow (a+2b)(b+2c)(c+2a)\geq 3(a+b+c)(ab+bc+ca) \]
\[ \Leftrightarrow ab^2+bc^2+ca^2\geq a^2b+b^2c+c^2a \]
\[ \Leftrightarrow (a-b)(b-c)(c-a)\geq 0 \]
上式成立，由于$c\geq b\geq a$
2、$a\geq b\geq c$
由AM-GM
\[  \frac{1}{[(a+b+c)+(2a+b)]^2}\leq \frac{1}{4(a+b+c)(2a+b)} \]
\[\Leftrightarrow \frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\leq \frac{a+b+c}{ab+bc+ca}\]
\[  \Leftrightarrow (2a+b)(2b+c)(2c+a)(a+b+c)\geq (ab+bc+ca)[2(a^2+b^2+c^2)+7(ab+bc+ca)]\]
由于$a\geq b\geq c$，所以
\[ (2a+b)(2b+c)(2c+a)\geq 3(a+b+c)(ab+bc+ca) \]
故只要证明
\[ 3(a+b+c)^2\geq 2(a^2+b^2+c^2)+7(ab+bc+ca) \]
显然成立。

证法2（西西）
\begin{align*}
 (a+2b+3c)^2&=[(a+c)+2(b+c)]^2\\
 &=(a+c)^2+4(b+c)^2+4(a+c)(b+c)\\
  &\geq 3(b+c)^2+6(a+c)(b+c)\\
  &=3(b+c)(2a+b+3c)
\end{align*}
等价证明
\[ \sum_{cyc}{\frac{1}{(b+c)(2a+b+3c)}}\leq \frac{3}{4(ab+bc+ca)}\]
就是
\[ \sum_{cyc}{\left(\frac{1}{2}-\frac{ab+bc+ca}{(b+c)(2a+b+3c)}\right)}\geq \frac{3}{4}\]
\[ \Leftrightarrow\sum_{cyc}{\frac{b+c}{2a+b+3c}}+2\sum_{cyc}{\frac{c^2}{(b+c)(2a+b+3c)}}\geq \frac{3}{2} \]
由Cauchy-Schwarz
\[ \sum_{cyc}{\frac{b+c}{2a+b+3c}}\geq \frac{4(a+b+c)^2}{\sum{(b+c)(2a+b+3c}}=1 \]
\[ 2\sum_{cyc}{\frac{c^2}{(b+c)(2a+b+3c)}}\geq \frac{2(a+b+c)^2}{\sum{(b+c)(2a+b+3c)}}=\frac{1}{2} \]
Done!

证法3（天书）
用替换
\[ \left\{
\begin{array}{ll}
x=3a+2b+c, &  \\
y=2b+2c+a, & \\
z=3c+2a+b, &
\end{array}\right.\]
得到\[ \left\{
\begin{array}{ll}
a=\dfrac{7x-5y+z}{18}, &  \\
b=\dfrac{x+7y-5z}{18}, & \\
c=\dfrac{y+7z-5x}{18}, &
\end{array}\right.\]
不等式等价于
\[ \Leftrightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\leq \frac{27}{14(xy+yz+xz)-11(x^2+y^2+z^2)} \]
\[ \Leftrightarrow 14\sum{xy}\cdot\sum{x^2y^2}\leq 27x^2y^2z^2+11\sum{x^2}\sum{x^2y^2} \]
\[ \Leftrightarrow 14\sum{x^3y^3}+14xyz\sum{x^2(y+z)}\leq 60x^2y^2z^2+11\sum{x^4(y^2+z^2)} \]
\[ \Leftrightarrow \sum{\left(7x^2y^2+2z^2(2x^2-3xy+2y^2)\right)(x-y)^2}\geq 0 \]
Done

问题：
设$a,b,c>0$，$a^2+b^2+c^2=3$证明
\[ \frac{(b+c)^2}{a^a+1}+\frac{(c+a)^2}{b^b+1}+\frac{(a+b)^2}{c^c+1}\leq 6 \]
证明:
注意到
\[ x^x\geq \frac{x^2+1}{2} \]
所以
\[ \sum{\frac{(b+c)^2}{a^a+1}}\leq \frac{2(b+c)^2}{a^2+3}=\sum{\frac{2(b+c)^2}{2a^2+b^2+c^2}}\]
再由Cauchy-Schwarz
\[ \sum{\frac{(b+c)^2}{2a^2+b^2+c^2}}\leq \sum{\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)}=3 \]
Done!

设$a,b,c,d>0$ 证明：
\begin{align*}
&\frac{9}{a(b+c+d)}+\frac{9}{b(c+d+a)}+\frac{9}{c(d+b+a)}+\frac{9}{d(a+b+c)}\\
&\geq\frac{16}{(a+b)(c+d)}+\frac{16}{(a+c)(b+d)}+\frac{16}{(a+d)(b+c)}
\end{align*}

（新加坡）
证明
两边同时乘以$a+b+c+d$,不等式变成
\begin{align}
 &\frac{9}{a}+\frac{9}{b}+\frac{9}{c}+\frac{9}{d}+\frac{9}{b+c+d}+\frac{9}{c+d+a}+\frac{9}{d+a+b}+\frac{9}{a+b+c}\\
 &\geq \frac{16}{a+b}+\frac{16}{c+d}+\frac{16}{a+c}+\frac{16}{b+d}+\frac{16}{a+d}+\frac{16}{b+c}
\end{align}
由Popoviciu不等式，我们有
\[ \frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{b+c+d}\geq \frac{4}{b+c}+\frac{4}{c+d}+\frac{4}{b+d}\]
所以有
\[ 3\sum_{sym}{\frac{1}{a}}+9\sum_{sym}{\frac{1}{a+b+c}}\geq 8\sum_{sym}{\frac{1}{a+b}}\]
故只要证明
\[ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq 3\sum_{sym}{\frac{1}{a+b+c}} \]
由AM-GM显然。$\square$