After subtitution:\(\displaystyle a=2\sin{\frac{A}{2}},b=2\sin{\frac{B}{2}},c=2\sin{\frac{C}{2}}\)
(where \(A,B,C\) is the 3 interior angles of a triangle)
the inequality is deduced to:
\[3+2(\cos A \cos B+\cos B \cos C+\cos C \cos A)\le 3(\cos A+\cos B+\cos C)\]
第一题 数学吧:http://tieba.baidu.com/f?kz=1143479157
如果非负实数\(a,b,c\)满足\(a^2+b^2+c^2+abc=4\) , 求证 :
\[a^2b^2+b^2c^2+c^2a^2+abc\le 4~.\]
After subtitution:\(\displaystyle a=2\sin{\frac{A}{2}},b=2\sin{\frac{B}{2}},c=2\sin{\frac{C}{2}}\)
(where \(A,B,C\) is the 3 interior angles of a triangle)
the inequality is deduced to:
\[3+2(\cos A \cos B+\cos B \cos C+\cos C \cos A)\le 3(\cos A+\cos B+\cos C)\]
By these formulas\[\cos A + \cos B + \cos C = \frac{{R + r}}{R}\]
\[\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{C}\cos{A}=\frac{s^2+r^2-4R^2}{4R^2}\]
the inequality is equivalent to\[s^2\le 4R^2+6Rr-r^2\]
which is true by \(\textrm{Gerretsen Inequality and Euler Inequality}\).第二题 数学吧 : http://tieba.baidu.com/p/586512170
实数\(a,b,c\)满足\(a^2+b^2+c^2=2\) , 求证 :
\[a+b+c\le 2+abc~.\]
By Cauchy-Schwartz inequality , we have
\[\begin{aligned}(a+b+c-abc)^2=&(a+b+(1-ab)c)^2\\ \le & ((a+b)^2+c^2)(1+(1-ab)^2)\\ =&2(1+ab)(2-2ab+a^2b^2)\end{aligned}\]
but\[4-2(1+ab)(2-2ab+a^2b^2)=a^2b^2(2-2ab)=a^2b^2((a-b)^2+c^2)\ge 0\]
done!第三题 数学吧 : http://tieba.baidu.com/p/1157372577
在\(\triangle ABC\)中 , 求证 :
\[(1-\cos A)(1-\cos B)(1-\cos C)\ge \cos A\cos B\cos C~.\]
证明:设\(\triangle ABC\)的三边长\(a,b,c\) , 则不等式等价于
\[\begin{aligned}&(a+b-c)^2(b+c-a)^2(c+a-b)^2\\ \ge &(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)\end{aligned}\]
如果\(a^2,b^2,c^2\)不能组成某三角形三边长 , 则\[(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2\le 0\]
此时不等式明显成立 .如果\(a^2,b^2,c^2\)能组成某三角形三边长 , 则设该三角形面积为\(S\) , 于是由费哈不等式有
\[2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4\ge 4\sqrt{3}S\]
,利用海伦公式即得等价于\[\begin{aligned}&(a+b+c)(a+b-c)(b+c-a)(c+a-b)\\ \ge &\sqrt{3(a^2+b^2+c^2)(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)}\end{aligned}\]
而因为\(a+b+c\le \sqrt{3(a^2+b^2+c^2)}\) , 所以\[\begin{aligned}&(a+b-c)(b+c-a)(c+a-b)\\ \ge &\sqrt{(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)}\end{aligned}\]
平方后即得\[\begin{aligned}&(a+b-c)^2(b+c-a)^2(c+a-b)^2\\ \ge &(a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)\end{aligned}\]
综上所述 , 不等式得证 .By these formulas
\[\cos A + \cos B + \cos C = \frac{{R + r}}{R}\]
\[\cos A\cos B\cos C = \frac{{s^2 - \left({r + 2R} \right)^2 }}{{4R^2 }}\]
\[\cos{A}\cos{B}+\cos{B}\cos{C}+\cos{C}\cos{A}=\frac{s^2+r^2-4R^2}{4R^2}\]
the inequality is equivalent to\[s^2\le 4R^2+4Rr+3r^2\]
which is \(\textrm{Gerretsen Inequality}\).第四题 数学吧 : http://tieba.baidu.com/p/1168567925
已知\(a_i>0~,~x_i\in[0,1]~,~i=1,2,\cdots,n\) , 且\(\displaystyle \sum_{i=1}^n a_i=1\) , 求证 :
\[\displaystyle \sum_{i=1}^n \frac{a_i}{1+x_i}\le \left(1+\prod_{i=1}^n x_i^{a_i}\right)^{-1}~.\]
证明 : 用Jensen不等式 , 构造函数\(\displaystyle f(y)=\frac{1}{1+e^y}\) , 然后设\(y_i=\ln x_i\) .
第五题 数学吧 : http://tieba.baidu.com/p/1163998974
已知正实数\(a,b,c\)满足\(2a+4b+7c\le 2abc\) , 求\(a+b+c\)的最小值 .
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=414104
第六题 数学吧 http://tieba.baidu.com/p/1163998974
正实数\(x,y,z\)满足\(x+y+z=3\) , 求证 :
\[\frac{x^3}{y^3+8}+\frac{y^3}{z^3+8}+\frac{z^3}{x^3+8}\ge \frac{1}{3}\ge \frac{1}{9}+\frac{2(xy+yz+zx)}{27}\]
第七题 数学吧 : http://tieba.baidu.com/p/1163998974
已知正实数\(a,b,c\)满足\(a+b+c=1\) , 求证 :
\[\sqrt{1+\frac{bc}{a}}+\sqrt{1+\frac{ca}{b}}+\sqrt{1+\frac{ab}{c}}\ge 2\sqrt{3}\]
第八题 数学吧 : http://tieba.baidu.com/p/1171523887
已知实数\(x_1,x_2,\cdots,x_n,y_1,y_2,\cdots,y_n\)满足\(\sum\limits_{i=1}^n x_i^2 \le1\) , 求证:
\[\left(\sum_{i=1}^n x_iy_i -1\right)^2\ge\left(\sum_{i=1}^n x_i^2 -1\right)\left(\sum_{i=1}^n y_i^2 -1\right) \]
证明1 : 如果\(\displaystyle\sum_{i=1}^n y_i^2\ge 1\) , 此时不等式左边非负 , 右边非正 , 不等式成立 .
如果\(\displaystyle\sum_{i=1}^n y_i^2\le 1\) , 由均值不等式 , 我们有
\[\begin{aligned}&\left(\sum_{i=1}^n x_i^2-1\right)\left(\sum_{i=1}^n y_i^2-1\right)\\ \le & \frac{1}{4}\left(\left(1-\sum\limits_{i=1}^n x_i^2\right)+\left(1-\sum\limits_{i=1}^n y_i^2\right)\right)^2\\ \le&\left(1-\sum_{i=1}^n x_iy_i\right)^2\end{aligned}\]
其中用到了\[0\le\frac{1}{2}\left(1-\sum\limits_{i=1}^n x_i^2+1-\sum\limits_{i=1}^n y_i^2\right)\le 1-\sum_{i=1}^n x_iy_i\]
证明2 : 如果\(\displaystyle\sum_{i=1}^n y_i^2\ge 1\) , 此时不等式左边非负 , 右边非正 , 不等式成立 .
如果\(\displaystyle\sum_{i=1}^n y_i^2\le 1\) , 设
\[x_{n+1}^2=1-\sum_{i=1}^n x_i^2\quad,\quad y_{n+1}^2=1-\sum_{i=1}^n y_i^2\]
其中\(x_{n+1},y_{n+1}\)非负 . 由柯西不等式 , 有\[1=\sqrt{\sum_{i=1}^{n+1} x_i^2\sum_{i=1}^{n+1} y_i^2}\ge\sum_{i=1}^n x_iy_i+x_{n+1}y_{n+1}\]
从而\[\left(1-\sum_{i=1}^n x_iy_i\right)^2\ge x_{n+1}^2y_{n+1}^2=\left(1-\sum_{i=1}^n x_i^2\right)\left(1-\sum_{i=1}^n y_i^2\right)\]
第九题 数学吧 http://tieba.baidu.com/p/1171951655
已知正实数\(a,b,c\)满足\(a+b+c=3\) , 求证:
\[\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge a^2+b^2+c^2\]
see also:
http://www.pep.com.cn/rjwk/gzsxsxkj/sxkj3/sxkj3ts/201105/t20110516_1041458.htm
http://www.pep.com.cn/rjwk/gzsxsxkj/sxkj4/sxkj4ts/201106/t20110623_1050799.htm
(题目3.1.9)
第十题 数学吧 http://tieba.baidu.com/p/1171951655
已知\(x,y,z>0\) , 求证 : \(\displaystyle\sqrt{1+\frac{48x}{y+z}}+\sqrt{1+\frac{48y}{z+x}}+\sqrt{1+\frac{48z}{x+y}}\ge 15\) .
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=64122