\(a,\)Ta có: \(\hept{\begin{cases}MA=MB\\OA=OB=R\end{cases}}\)

\(\Rightarrow MO\)là đường trung trực của \(AB\)

   \(\Rightarrow MO\perp AB\)tại trung điểm \(K\)của \(AB\)        

\(b,\)Áp dụng hệ thức lượng vào tam giác vuông \(MAO\)có:

\(+\)\(^{^{ }OA^2+AM^2=OM^2\Leftrightarrow AM=\sqrt{OM^2-OA^2}\Leftrightarrow AM=\sqrt{\frac{8}{5}R)^2-R^2}\Leftrightarrow AM=\frac{\sqrt{39}R}{5}}\)

\(+\) \(AK.OM=OA.AM\Leftrightarrow AK.\frac{8}{5}R\)\(=R.\frac{\sqrt{39}}{5}R\Rightarrow AB=2AK=R\frac{\sqrt{39}}{4}\)

\(+\) \(OA^2=OK.ON\Leftrightarrow OK=\frac{OA^2}{ON}=\frac{R^2}{\frac{8R}{5}}\)\(=\frac{5R}{8}\)

\(c,\)Ta có: \(\widehat{ABN}=90\)(B thuộc đường tròn đường kính AN) \(\Rightarrow BN//MO\left(\perp AB\right)\)

Do đó; \(\hept{\begin{cases}\widehat{AOM=\widehat{ANB}}\\\widehat{AOM=\widehat{BOM}}\end{cases}}\)

\(\Rightarrow\widehat{BOM=\widehat{ANB}}\)

Xét tam giác BHA  và MBO có:

\(\hept{\begin{cases}\widehat{BHN}=\widehat{MBO}=90\\\widehat{BNH}=\widehat{BOM}\end{cases}}\)\(\Rightarrow\Delta BHN\simeq\Delta MBO\)\(\Rightarrow\hept{\begin{cases}BH=BN\\MB=MO\end{cases}}\)\(\Rightarrow BH.MO=BN.MB\left(đpcm\right)\)

Đặt \(\frac{ab}{c}=x;\frac{bc}{a}=y;\frac{ca}{b}=z\Rightarrow xy=b^2;yz=c^2;xz=a^2\)

Ta có : \(\hept{\begin{cases}\left(x-y\right)^2\ge o\\\left(y-z\right)^2\ge0\\\left(x-z\right)^2\ge0\end{cases}}\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)\ge3\left(xy+yz+xz\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

\(\Leftrightarrow\sqrt{\left(x+y+z\right)^2}\ge\sqrt{3\left(xy+yz+xz\right)}\)

\(\Leftrightarrow\sqrt{\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)^2}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)( a,b,c là số thực dương ) ( ĐPCM )

bạn dung bđt a+b >= 2 căn ab ( cô si )  nhé

cách là ghép từng cặp ở vế trái lại

 Ta có: \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\)

\(=\frac{1}{2}\left(\frac{ab}{c}+\frac{bc}{a}\right)+\frac{1}{2}\left(\frac{bc}{a}+\frac{ca}{b}\right)+\frac{1}{2}\left(\frac{ca}{b}+\frac{ab}{c}\right)\)

\(\ge\frac{1}{2}\cdot2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}+\frac{1}{2}\cdot2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}+\frac{1}{2}\cdot2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}\) (Cauchy)

\(=\frac{1}{2}\cdot2b+\frac{1}{2}\cdot2c+\frac{1}{2}\cdot2a\)

\(=a+b+c\)

Dấu "=" xảy ra khi: a = b = c

\(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}=\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}}{\sqrt{a}+\sqrt{b}}=\)

\(=\frac{a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a}+\sqrt{b}}=\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left[\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\right]}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\)

\(=\left(\sqrt{a}-\sqrt{b}\right)^2\left(dpcm\right)\)