|
|
Author |
hbghlyj
Posted 2019-12-5 18:30
Last edited by hbghlyj 2019-12-29 23:46第三组.三角不等式
(1)求k的最小值,使得$(\cos A-\frac15)(\cos B-\frac15)(\cos C-\frac15)\leq k$对任意△ABC均成立
(2)三角形中,求$\csc\frac A2\csc\frac B2(4\sqrt{1+5\csc\frac C2})$的取值范围
(3)$\frac{\tan^2x+\tan^2y}{1+\tan^2x+\tan^2y}=\sin^2x+\sin^2y$,求$\sin x\sin y$的取值范围
切化弦,$\sin^2x\sin^2y=1-\cos^4x\in[0,1)$,原式当x=0,y为除$\frac{(2k+1)\pi}2$外的任意实数时取最小值,当x$\to\frac{(2k+1)\pi}2,y\to\frac{(2k+1)\pi}2$时取最大值.
(4)在锐角三角形中求$\frac{\sum\frac{\cos^2A}{\cos^2B}}{\sum\cos^2A}$的取值范围
$∵\prod\cos A≤\frac18∴\frac{2\cos^2A}{\cos^2B}+\frac{\cos^2B}{\cos^2C}\geq3\sqrt[3]{\frac{\cos^2A\cos^2A\cos^2B}{\cos^2B\cos^2B\cos^2C}}=\frac{3\cos^2A}{\sqrt[3]{\cos^2A\cos^2B\cos^2C}}≥12\cos^2A$,轮换求和,$\frac{\sum\frac{\cos^2A}{\cos^2B}}{\sum\cos^2A}≥4$,当且仅当$A=B=C=\frac{\pi}3$时取等
当A$\to\frac{\pi}2$时原式$\to\infty$,所以取值范围为[4,∞)
(5)$x,y,z>0,2\prod(1-x)=(1-x-y-z)^2,$求$\frac{\sum\frac{1-x}{1-y}}{1-2xyz}$的取值范围
使用恒等式$x^2+y^2+z^2+2xyz=1$将x换成$-\cos2A$就得到(4).
这样改编应该没问题。机器算出min{((1 - x)/(1 - y) + (1 - y)/(1 - z) + (1 - z)/(1 - x))/(1 - 2 x y z)|x^2 + 2 x y z + y^2 + z^2 = 1 ∧ x>y>z>0}≈4. at (x, y, z)≈(0.5, 0.5, 0.5)
max{((1 - x)/(1 - y) + (1 - y)/(1 - z) + (1 - z)/(1 - x))/(1 - 2 x y z)|x^2 + 2 x y z + y^2 + z^2 = 1 ∧ x>y>z>0}≈36.7174 at (x, y, z)≈(0.972157, 0.234333, 0)
请问这个最大值是怎么回事
(6)x,y,z>0,证明$\sum\frac x{x+\sqrt{(x+y)(x+z)}}≤1$
由齐次性,不妨设$xy+yz+zx=1$,$LHS=\sum\frac 1{1+\sqrt{1+x^{-2}}}(令x=\tan \frac A2)=\sum\frac{\sin\frac A2}{\sin\frac A2+1}=3-\sum\frac1{\sin\frac A2+1}≤3-\frac9{\sum\sin\frac A2+3}≤1$,当且仅当x=y=z时取等
(7)a,b,c,d>0,$\sum\frac{a^2}{a^2+1}=1,$求abcd的取值范围
令$a=\tan A$,其余类推,$A,B,C,D\in(0,\frac{\pi}2),\sum\sin^2 A=1,\sin^2 A+\sin^2 B+\sin^2 C=\cos^2 D≥3\sqrt[3]{\sin^2 A\sin^2 B\sin^2 C}$,轮换相乘得$\prod\tan^2 A≤\frac1{81},\prod\tan A≤\frac1{9}$,当且仅当a=b=c=d=$\sqrt3$时取等
(8)a,b,c>0,abc+a+c=b,求$P=\frac{2}{a^2+1}-\frac{2}{b^2+1}+\frac{3}{c^2+1}$
设$a=\tan x,b=\tan y,c=\tan z,x,y,z\in(0,\frac{\pi}{2})$,则$\tan y=\frac{\tan x+\tan z}{1-\tan x\tan z},y=x+z,P=2\cos^2 x-2\cos^2y+3\cos ^2z=2\sin (2x+z)\sin z-3\sin ^2 z+3≤2\sin z-3\sin ^2 z+3≤\frac{10}{3}$当且仅当$\sin x=\frac1{\sqrt3},\sin z=\frac13,a=\frac{\sqrt2}2,b=\sqrt2,c=\frac{\sqrt2}4$时取等
(9)解不等式min{sinx,cosx}<min{tanx,cotx}
$(k\pi,(k+\frac12)\pi)$作图
(10)任给7个实数存在两个a,b满足$0≤\frac{a-b}{1-ab}≤\frac{\sqrt3}3$
令$a=\tan x,b=\tan y,x,y\in(-\frac{\pi}2,\frac{\pi}2),$则$0≤\tan (x-y)≤\frac{\sqrt3}3,0≤x-y≤\frac{\pi}6$,把$(-\frac{\pi}2,\frac{\pi}2)$等分为6个区间,每个区间长度为$\frac{\pi}6$,由抽屉原理得证
(11)
常用三角形三内角不等式
(1)$\sum\cos A≤\frac32$
$\sum\cos A=2\cos{\frac{A+B}2}\cos{\frac{A-B}2}+1-2\sin^2\frac C2(消去自由变量\cos{\frac{A-B}2})≤2\sin\frac C2+1-2\sin^2\frac C2≤\frac32$
(2)$\prod\cos A≤\frac18$
令$x=\cot A,y=\cot B,z=\cot C$,则$\sum xy=1,\prod\cos A=\frac{xyz}{\sqrt{\prod(1+x^2)}}=\frac{xyz}{\prod(x+y)}≤\frac{xyz}{\prod2\sqrt{xy}}=\frac18$
(3)$\prod\sin\frac A2≤\frac18$
$\prod\sin\frac A2=\frac12(\cos{\frac{A+B}2}-\cos{\frac{A-B}2})\sin\frac C2(消去自由变量\cos{\frac{A-B}2})≤\frac12(1-\sin C)\sin\frac C2≤\frac18$
函数角度上(2)(3)是等价的
(1)可以通过$\sum\cos A=4\prod\sin\frac{A}2+1$导出(3)
上面三个都可用Jensen不等式做 |
|
kuing
Posted 2019-11-29 22:33
其妙
Posted 2020-1-14 23:23
