来自天书的不等式证明

Eufisky posted @ 2014年8月25日 02:33 in 不等式 with tags 天书不等式 , 1273 阅读

(a)$a,b,c\geq0$，则下式成立当且仅当参数$k\geq\frac{\sqrt{5}-1}{4}$，有$\left( {\frac{a}{{b + c}} + k} \right)\left( {\frac{b}{{c + a}} + k} \right)\left( {\frac{c}{{a + b}} + k} \right) \ge {\left( {\frac{1}{2} + k} \right)^3}.$

(b)$a,b,c,d\geq0$，则下式成立当且仅当参数$k\geq k_0$，$\left( {\frac{a}{{b + c + d}} + k} \right)\left( {\frac{b}{{c + d + a}} + k} \right)\left( {\frac{c}{{d + a + b}} + k} \right)\left( {\frac{d}{{a + b + c}} + k} \right) \ge {\left( {\frac{1}{2} + k} \right)^4}.$其中：${k_0} = \frac{{2\sqrt 6 }}{9}\cos \left( {\frac{1}{3}\arccos \frac{{27\sqrt 6 }}{{64}}} \right) - \frac{1}{6} \approx 0.379.$

$$108k^3+54k^2-15k-8\ge0$$

$$k_0=\frac{2\sqrt{6}}{9}\cos\left(\frac{1}{3}\arccos\left(\frac{27\sqrt{6}}{64}\right)\right)-\frac{1}{6}$$

$$\left\{\begin{array}{cc}3a-xb-xc-xd=0\\-ya+3b-yc-yd=0\\-za-zb+3c-zd=0\\-wa-wb-wc+3d=0\end{array}\right.$$

$$\left|\begin{array}{cccc}3&-x&-x&-x\\-y&3&-y&-y\\-z&-z&3&-z\\-w&-w&-w&3\end{array}\right|=0\Rightarrow 3\sum_{sym}xy+2\sum xyz+xyzw=27$$

$$\prod(3k+x)\ge(3k+1)^4\Leftrightarrow xyzw+3k\sum xyz+9k^2\sum_{sym}xy+27k^3\sum x\ge(3k+1)^4-(3k)^4$$

$$(3k-2)\sum xyz+(9k^2-3)\sum_{sym}xy+27k^3\sum x\ge(3k+1)^4-(3k)^4-27$$

$$\sum x\ge4;\sum_{sym}xy\ge6;3\sum x\ge2\sum_{sym}xy;\sum xyz\le4;xyzw\le1$$

\begin{align*}&(3k-2)\sum xyz+(9k^2-3)\sum_{sym}xy+27k^3\sum x\\&\ge(9k^2-3-\frac{3}{2}(3k-2))\sum_{sym}xy-13(3k-2)+27k^3\sum x\\&\ge(9k^2-\frac{9k}{2})\sum_{sym}xy-13(3k-2)+18k^3\sum_{sym}xy\end{align*}

$$(18k^3+9k^2-\frac{9k}{2})\sum_{sym}xy\ge13(3k-2)+(3k+1)^4-(3k)^4-27$$
$$\Leftrightarrow k(4k^2+2k-1)\left(\sum_{sym}xy-6\right)\ge0$$

$$4k^2+2k>4(\frac{7}{20})^2+2(\frac{7}{20})>1$$

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