câu hình:
a) Vì C là điểm chính giữa cung AB \(\Rightarrow OC\bot AB\Rightarrow\angle AOC=90\)
\(\Rightarrow\angle AOC=\angle AHC\Rightarrow AOHC\) nội tiếp
b) Vì AOHC nội tiếp \(\Rightarrow\angle CHO=180-\angle CAO=180-\angle CAB=\angle CNB\)(CANB nội tiếp)
c) Xét \(\Delta CHM\) và \(\Delta ACM:\) Ta có: \(\left\{{}\begin{matrix}\angle CHM=\angle ACM=90\\\angle CMAchung\end{matrix}\right.\)
\(\Rightarrow\Delta CHM\sim\Delta ACM\left(g-g\right)\Rightarrow\dfrac{HM}{CM}=\dfrac{CM}{MA}\)
Xét \(\Delta BNM\) và \(\Delta ACM:\) Ta có: \(\left\{{}\begin{matrix}\angle BMN=\angle AMC\\\angle CAM=\angle MBN\left(ACNBnt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta BNM\sim\Delta ACM\left(g-g\right)\Rightarrow\dfrac{MN}{BM}=\dfrac{CM}{MA}\)
\(\Rightarrow\dfrac{MN}{BM}=\dfrac{MH}{CM}\) mà \(BM=CM\Rightarrow MH=MN\)
\(\Rightarrow BHCN\) là hình bình hành (2 đường chéo giao nhau tại trung điểm mỗi đường)
\(\Rightarrow\angle IHB=\angle ICN=90-\angle CNA=90-\angle CBA=45\) (C là điểm chính giữa)
mà \(\angle IHO=\angle CAO=45\Rightarrow\angle OHB=90\Rightarrow OH\bot HB\)
Ta có: \(CH^2=AH.HM\Rightarrow AH=\dfrac{CH^2}{HM}=\dfrac{NB^2}{\dfrac{1}{2}HN}=\dfrac{2BN^2}{HN}\)
Lại có: \(\angle NHB=90-\angle BHI=90-45=45\Rightarrow\Delta NHB\) vuông cân
\(\Rightarrow BN=HN\Rightarrow AH=\dfrac{2BN^2}{BN}=2BN=BN+HN\)
d) Vì \(\angle OHI=\angle BHI=45\Rightarrow HI\) là phân giác \(\angle OHB\)
\(\Rightarrow\dfrac{IO}{IB}=\dfrac{OH}{HB}\)
Xét \(\Delta OHB\) và \(\Delta CHA:\) Ta có: \(\left\{{}\begin{matrix}\angle CHA=\angle OHB=90\\\angle ACH=\angle HOB\end{matrix}\right.\)
\(\Rightarrow\Delta OHB\sim\Delta CHA\left(g-g\right)\Rightarrow\dfrac{OH}{HB}=\dfrac{CH}{AH}=\dfrac{BN}{BN+HN}=\dfrac{BN}{2BN}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{IO}{IB}=\dfrac{1}{2}\Rightarrow IB=2IO\)
câu 5 ta có: \(2021\left(x^2+y^2+z^2\right)=3xyz\)
\(=>\dfrac{x^2+y^2+z^2}{xyz}=\dfrac{3}{2021}< =>\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=\dfrac{3}{2021}\)
Áp dụng BDT Cô si
\(=>\left\{{}\begin{matrix}\dfrac{x}{yz}+\dfrac{y}{xz}\ge\dfrac{2}{z}\\\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{2}{x}\\\dfrac{x}{yz}+\dfrac{z}{xy}\ge\dfrac{2}{y}\end{matrix}\right.\)\(\)
\(=>\left(\dfrac{x}{yz}+\dfrac{y}{xz}\right)+\left(\dfrac{y}{xz}+\dfrac{z}{xy}\right)+\left(\dfrac{x}{yz}+\dfrac{z}{xy}\right)\ge2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(=>\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
\(=>\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{3}{2021}\)
Áp dụng cố si \(=>x^2+yz\ge2x\sqrt{yz}=>\dfrac{x}{x^2+yz}\le\dfrac{1}{2\sqrt{yz}}=\dfrac{1}{4}.2.\dfrac{1}{\sqrt{y}}.\dfrac{1}{\sqrt{z}}\)\(=\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)\)(1)
tương tự \(=>\dfrac{y}{y^2+zx}\le\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)\left(2\right)\)
\(\dfrac{z}{z^2+xy}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(3\right)\)
cộng vế (1)(2)(3)
\(=>A\le\dfrac{1}{4}\left[\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right]\)\(=\dfrac{1}{4}.2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(=\dfrac{1}{2}.\dfrac{3}{2021}=\dfrac{3}{4042}\). Dấu"=" xảy ra<=>\(x=y=z=\dfrac{1}{2021}\)
vậy Max \(=\dfrac{3}{4042}\)
