\(a^3+b^3+1=a^3+b^3+abc\ge ab\left(a+b+c\right)\)

=>  \(\frac{\sqrt{1+a^3+b^3}}{ab}\ge\frac{\sqrt{ab\left(a+b+c\right)}}{ab}=\frac{\sqrt{a+b+c}}{\sqrt{ab}}\)

Tuong tu:  \(\frac{\sqrt{1+b^3+c^3}}{bc}\ge\frac{\sqrt{a+b+c}}{\sqrt{bc}}\)

                    \(\sqrt{1+c^3+a^3}\ge\frac{\sqrt{a+b+c}}{\sqrt{ca}}\)

suy ra:  \(\frac{\sqrt{1+a^3+b^3}}{ab}+\frac{\sqrt{1+b^3+c^3}}{bc}+\frac{\sqrt{1+c^3+a^3}}{ca}\ge\sqrt{a+b+c}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)\)

\(\ge\sqrt{3\sqrt[3]{abc}}.3\sqrt[3]{\frac{1}{\sqrt{ab}}.\frac{1}{\sqrt{bc}}.\frac{1}{\sqrt{ca}}}=3\sqrt{3}\)  (dpcm)

a: Xét (O) co

CM,CA là tiếp tuyên

=>CM=CA 

Xét (O) có

DM,DB là tiếp tuyến

=>DM=DB

CD=CM+MD

=>CD=CA+BD

b: Xet ΔACN và ΔDBN có

góc NAC=góc NDB

góc ANC=góc DNB

=>ΔACN đồng dạng vơi ΔDBN

=>AC/BD=AN/DN

=>CN/MD=AN/ND

=>MN//AC//BD

C/m: BDT:  \(a^3+b^3+abc\ge ab\left(a+b+c\right)\)   (1)

That vay ta co:

\(a^3+b^3+abc-ab\left(a+b+c\right)=\left(a+b\right)\left(a-b\right)^2\ge0\)   (luon dung)

Tuong tu ta co:  \(b^3+c^3+abc\ge bc\left(a+b+c\right)\)  (2)

                         \(c^3+a^3+abc\ge ca\left(a+b+c\right)\)   (3)

Tu (1), (2), (3)  suy ra:

\(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)

\(=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}\)   (dpcm)

\(Taco:\)

\(x^2\ge0\forall x\Rightarrow2x^2\ge0\)

\(\left(+\right)x>0\Rightarrow2x^2\ge2\Rightarrow2x^2+x-1\ge2\left(loại\right)\)

\(\left(+\right)x< 0\Rightarrow2x^2+x-1\ge0\left(loại\right)\)

\(Vậy:x=0.Taco:2x^2+x-1=-1< 0\left(thoaman\right)\)

\(2x^2+x-1< 0\)

<=>  \(\left(x+1\right)\left(2x-1\right)< 0\)

TH1:   \(\hept{\begin{cases}x+1>0\\2x-1>0\end{cases}}\)<=>    \(\hept{\begin{cases}x>-1\\x>\frac{1}{2}\end{cases}}\) <=>  \(x>\frac{1}{2}\)

TH2:  \(\hept{\begin{cases}x+1< 0\\2x-1< 0\end{cases}}\) <=>  \(\hept{\begin{cases}x< -1\\x< \frac{1}{2}\end{cases}}\)  <=>  \(x< -1\)

Vay....

tu ve hinh nhe:

a)  Theo t/c 2 tiep tuyen cat nhau ta co:  OA la phan giac goc BOC

Tam giab BOC can tai O, OA la phan giac goc BOC

=> OA là đường trung trực của BC

hay OA vuong goc voi BC

Ap dung HTL vao tam giac vuong ABO ta co:

               \(OH.OA=OB^2=R^2\)    (dpcm)

b)  De thay:  tam giac BFE vuong tai F

hay BF vuong goc voi AE

Ap dung HTL vao 2 tam giac vuong: ABO va BAE ta co:

    \(AH.AO=AB^2\)

    \(AF.AE=AB^2\)

suy ra:   \(AH.AO=AF.AE\)

c)  tu b) c/m:  \(\Delta AHF~\Delta AEO\) (c.g.c)

=>   \(\widehat{AHF}=\widehat{AEO}\)

Ta co:   \(\widehat{AHF}+\widehat{OHF}=180^0\)

=>   \(\widehat{AEO}+\widehat{OHF}=180^0\)