Aug 28

## 一个不等式（百度贴吧）

$\sqrt{\frac{a}{a^2+b^2+1}}+\sqrt{\frac{b}{b^2+c^2+1}}+\sqrt{\frac{c}{c^2+a^2+1}}\leq \sqrt{3}$

$\left(\sum\sqrt{\frac{a}{a^2+b^2+1}}\right)^{2}\leq \sum(a^2+c^2+1)\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}=9\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}$

$\sum \frac{a}{(a^2+b^2+1)(a^2+c^2+1)}\leq \frac{1}{3}$
$\Leftrightarrow 12\sum{a}-3\sum{a^3}\leq 34-a^2b^2c^2-2\sum{a^4}$

$12\sum{a}-3\sum{a^3}\leq 33-2\sum{a^4}$
$\Leftrightarrow \sum_{cyc}(1-a^2)\left(2a^3-3a+2+\frac{9}{a+1}\right)\geq 0$

$\sum(1-a^2)S_{a}\geq \frac{1}{3}\sum(1-a^2)\sum S_{a}=0$
Done!

Aug 26

## Tourish 的一个根式不等式

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 $a=b=c=1$.

Aug 25

## 来自西西解疑群中的一个不等式

Problem

$\sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}$

Solution

Cauchy-Schwarz

$[(a^2+1)+20](5+20)\geq \left[\sqrt{5(a^2+1)}+20\right]^{2}$
$[ 2(b^2+1)+12](4+12)\geq \left[\sqrt{8(b^2+1)}+12\right]^2$
$[(c^2+1)+90](10+90)\geq \left[\sqrt{10(c^2+1)}+90\right]^2$

$\sqrt{a^2+21}\geq \sqrt{\frac{1}{5}\left(a^2+1\right)}+4$
$\sqrt{2b^2+14}\geq \sqrt{\frac{1}{2}\left(b^2+1\right)}+3$
$\sqrt{c^2+91} \geq \sqrt{\frac{1}{10}\left(c^2+1\right)}+9$
\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*}

$(a^2+1)(b^2+1)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2\geq (ab+bc+ca-1)^2\ge 100$

$16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}\geq 19$
$\sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}\geq 19$

Jul 21

## quyhktn-qa1的2个不等式

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!          $\square$

Jun 22

## AOPS上的一个带绝对值的不等式

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
$\frac{2}{a-b}+\frac{2}{b-c}\geq \frac{8}{a-c}$
$P\geq \frac{10}{a-c}+\frac{5}{\sqrt{ab+bc+ca}}\geq \frac{4}{\frac{a-c}{10}+\frac{\sqrt{ab+bc+ca}}{5}}$
$\frac{a-c}{10}+\frac{\sqrt{ab+bc+ca}}{5}\leq \frac{4}{10\sqrt{6}}=\frac{1}{15}\sqrt{6}$
$(a-c)+2\sqrt{(a+c)-(a+c)^2+ac}\leq \frac{2}{3}\sqrt{6}$