陈洪葛的博客

设$x,y,z\in R$,证明

\[ 5(x^4+y^4+z^4)+7(x^3y+y^3z+z^3x)\geq 0 \]
证明：
仿照以前的做法我们可以把不等式加强为有基本取等条件$x=y=z$的那种。
\[ 5(x^4+y^4+z^4)+7(x^3y+y^3z+z^3x)\geq \frac{36}{81}(x+y+z)^4 \]
展开（这个系数有点大，不着急，慢慢展）
\[ 369\sum{x^4}+423\sum{x^3y}-143\sum{xy^3}-216\sum{x^2y^2}-432\sum{x^2yz}\geq 0 \]
到至今为止，这个3元4次不等式已经不是问题了，直接套Vo Quoc Ba Can配方文章中的结论，
只要验证
\[ \left\{
\begin{array}{ll}
 m>0, & \\
 3m(m+n)\geq p^2+pg+g^2, &
 \end{array}
   \right.\]
这里
\[ m=369,n=-216, p=423,g=-143 \]
\[ 3m(m+n)-(p^2+pg+g^2)=169371-(178929+20449-60189)=30182>0 \]
所以不等式得证。                      $\blacksquare$

同样的手段奏效于

\[ 3(x^4+y^4+z^4)+4(x^3y+y^3z+z^3x)\geq 0  \]

事实上，我们可以用pqr来弄出那个最佳系数 :)

\[ (x^4+y^4+z^4)+k(x^3y+y^3z+z^3x)\geq 0 \]

Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that
\[ \sqrt[3]{3a^2+2ab+3b^2}+\sqrt[3]{3b^2+2bc+3c^2}+\sqrt[3]{3c^2+2ca+3a^2} \geq 6.\]

Proof：(by gxggs)

By Holder inequality, we have
\[\left(\sum_{cyc}{\sqrt[3]{3a^2+2ab+3b^2}}\right)^3\sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\geq 16(a+b+c)^4\]
Therefore it suffices to show that
\[ 16(a+b+c)^4\geq 216\cdot \sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\]
Or
\[2(a+b+c)^2\geq 3\cdot \sum_{cyc}{\left(\frac{(a+b)^4}{3a^2+2ab+3b^2}\right)}\]
\[\Longleftrightarrow \sum_{cyc}{\left(a^2+b^2+4ab-\frac{3(a+b)^4}{3a^2+2ab+3b^2}\right)}\geq 0\]
which is equivalent to
\[\frac{2ab(a-b)^2}{3a^2+2ab+3b^2}+\frac{2bc(b-c)^2}{3b^2+2bc+3c^2}+\frac{2ca(c-a)^2}{3c^2+2ca+3a^2}\geq 0\]
which is obvious true. Equal holds when $a=b=c=1$ or $a=3,b=0,c=0$ or their cyclic forms.   $\blacksquare$
另外，越南TaHongQuang 作了如下推广

 1.Let $a, \, b, \, c > 0$ such that $a+b+c =\frac{1}{4}$ .  Prove that
\[  \sqrt [3]{65{a}^{2}+81{b}^{2}-2ab}+\sqrt [3]{65{b}^{2}+81{c}^{2}-2bc}+\sqrt [3]{65{c}^{2}+81{a}^{2}-2ac} \ge 3 \]

2.Let $a, \, b, \, c > 0$ such that $a+b+c =\frac{1}{6}$ .  Prove that
\[  \sqrt [3]{7{a}^{2}+9{b}^{2}-4ba}+\sqrt [3]{7{b}^{2}+9{c}^{2}-4bc}+\sqrt [3]{7{c}^{2}+9{a}^{2}-4ac}\ge 1 \]

If $a,b,c\ge 0$ such that $ab+bc+ca=3$, then

$$4(a^3+b^3+c^3)+7abc+125\ge 48(a+b+c).$$

Proof(Vo Quoc Ba Can)

The inequality is equivalent to \[4(a^3+b^3+c^3)+7abc +125 \ge 16(ab+bc+ca)(a+b+c),\]

or

\[125 \ge A,\]

where $A=16(a+b+c)(ab+bc+ca)-4(a^3+b^3+c^3) -7abc.$

From the equivalence, it is obvious that the inequality holds for $A \le 0,$ so we only have to consider it when $A >0.$ In that case, it is equivalent to \[125^2 \ge A^2,\] and there are two possible cases can occur (assume that $a \ge b \ge c$):
1、Case $6(a+c) \ge 7b.$

Using the AM-GM inequality, we have

\[A^2 =(2a+2c+b)^2\cdot \frac{A}{2a+2c+b}\cdot \frac{A}{2a+2c+b} \le \left[\frac{(2a+2c+b)^2+2\cdot \frac{A}{2a+2c+b}}{3}\right]^3.\]

Thus, it suffices to prove that

\[25\cdot 3 \ge (2a+2c+b)^2+\frac{2A}{2a+2c+b},\]

or

\[25(ab+bc+ca) \ge (2a+2c+b)^2+\frac{2A}{2a+2c+b}.\]

An easy computation shows that, for the last inequality, we have

\[LHS-RHS =\frac{(a-b)(b-c)(6a+6c-7b)}{2a+2c+b},\]

which is obviously nonnegative.
2、Case $6(a+c) \le 7b.$

Applying the AM-GM inequality in the same way as the first case, but this time we change the positions of $a$ and $b$ ($a\to b$ and $b\to a$), it leads us to the inequality

\[25(ab+bc+ca) \ge (2b+2c+a)^2+\frac{2A}{2b+2c+a},\]

which is equivalent to

\[\frac{(a-b)(a-c)(7a-6b-6c)}{2b+2c+a} \ge 0.\]

This is true because

$$(a-b)(a-c) \ge 0$$

and

\[7a-6b-6c=13a-6(a+b+c) \ge 13b-6(a+b+c)=7b-(6a+6c) \ge 0.\]
The proof is completed. $\blacksquare$

Let $a,b,c>0$,with $a+b+c=3$,Prove that
\[ \sqrt{\frac{a^{3}}{a^{2}+3b^{2}}}+\sqrt{\frac{b^{3}}{b^{2}+3c^{2}}}+\sqrt{\frac{c^{3}}{c^{2}+3a^{2}}}\geq \frac{3}{2}\]

Proof:
\begin{align*}
\sum_{cyc}{\sqrt{\frac{a^{3}}{a^{2}+3b^{2}}}}&=6\sum_{cyc}{\frac{a^{2}}{\sqrt{4a(a+b+c)\cdot3(a^2+3b^2)}}}\\
&\geq 12\sum_{cyc}{\frac{a^2}{4a(a+b+c)+3(a^2+3b^2)}}\\
&=12\sum_{cyc}{\frac{a^{2}}{7a^{2}+9b^2+4ab+4ca}} 
\end{align*}
Now,Using the Awesome CYH technology,we have
\begin{align*}
&\left(\sum_{cyc}{\frac{a^{2}}{7a^{2}+9b^2+4ab+4ca}}\right)\left[\sum_{cyc}{(2a+c)^2(7a^2+9b^2+4ab+4ca)}\right]\\
&\geq \left[\sum_{cyc}{a(c+2a)}\right]^{2}\\
&=\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2 
\end{align*}
Thus,it's suffice to check
\[ 8\left(2\sum_{cyc}{a^2}+\sum_{cyc}{ab}\right)^2\geq \sum_{cyc}{(2a+c)^2(7a^2+9b^2+4ab+4ca)}\]
Or
\[ \sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2}+3\sum_{cyc}{a^3b}-3\sum_{cyc}{ab^3}-2abc\sum_{cyc}{a}\geq 0 \]
Which is not a problem for us now,so we are done!
$\square$

Let$a,b,c>0$ with $a + b+c=3$, Prove that $a\sqrt[3]{a+b}+b\sqrt[3]{b+c}+c\sqrt[3]{c+a} \ge 3 \sqrt[3]2$

(Tran Quoc Anh)

Proof:

by AM-GM inequality,we have
$$ a\sqrt[3]{a+b}=\frac{3\sqrt[3]{2}a(a+b)}{3\sqrt[3]{2(a+b)(a+b)}}\geq 3\sqrt[3]{2}\cdot \frac{a(a+b)}{2+2a+2b} $$
Thus,it's suffice to prove that
$$ \frac{a(a+b)}{a+b+1}+\frac{b(b+c)}{b+c+1}+\frac{c(c+a)}{c+a+1}\geq 2 $$
Or
$$ \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\leq 1 $$
After homogenous,it's
$$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\leq \frac{1}{3} $$
Now,multiply $4a+4b+4c$ to each sides.we can rewrite the inequality into
$$ \frac{9ca}{4a+4b+c}+\frac{9ab}{4b+4c+a}+\frac{9bc}{4c+4a+b}\leq a+b+c $$
Using Cauchy-Schwarz inequality,we have
$$ \frac{9}{4a+4b+c}=\frac{(2+1)^2}{2(2a+b)+(2b+c)}\le \frac{2}{2a+b}+\frac{1}{2b+c} $$
Therefore
\begin{align}
  \sum{\frac{9ca}{4a+4b+c}}&\leq \sum{\left(\frac{2ca}{2a+b}+\frac{ca}{2b+c}\right)}\\
&=a+b+c
\end{align}
Hence we are done!

未改进的结果

http://kkkkuingggg.5d6d.net/thread-1119-1-3.html