Áp dụng bất đẳng thức Bunhiacopxki, ta có:

\(\left(a+b+c\right)^2\le\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)=3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow1\le3\left(a^2+b^2+c^2\right)\Leftrightarrow a^2+b^2+c^2\ge\frac{1}{3}\)

Dấu "=" khi a=b=c

Từ\(\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab\)

\(\frac{a^2+c^2}{b^2+c^2}=\frac{a^2+ab}{b^2+ab}=\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\)

a, - Xét \(\Delta IKC\) và \(\Delta ICB\) có :

\(\left\{{}\begin{matrix}\widehat{KIC}\left(chung\right)\\\widehat{ICK}=\widehat{IBC}\left(=\frac{1}{2}Sđ\stackrel\frown{KC}\right)\end{matrix}\right.\)

=> \(\Delta IKC\) ~ \(\Delta ICB\) ( g - g )

=> \(\frac{IK}{IC}=\frac{IC}{IB}\)

=> \(IC^2=IK.IB\left(đpcm\right)\)

b, Ta có : BD // AC .

=> \(\widehat{BDA}=\widehat{DAC}\) ( So le trong )

Mà \(\widehat{BDA}=\widehat{ABI}\left(=\frac{1}{2}Sđ\stackrel\frown{BK}\right)\)

=> \(\widehat{DAC}=\widehat{ABI}\) .

- Xét \(\Delta AIK\) và \(\Delta BIA\) có :

\(\left\{{}\begin{matrix}\widehat{AIB}\left(chung\right)\\\widehat{DAI}=\widehat{ABI}\left(cmt\right)\end{matrix}\right.\)

=> \(\Delta AIK\) ~ \(\Delta BIA\) ( g - g )

=> \(\frac{AI}{IK}=\frac{IB}{AI}\)

=> \(AI^2=IK.IB\)

Mà \(IC^2=IK.IB\) ( câu a )

=> \(AI=IC\left(đpcm\right)\)

c, not hiểu câu hỏi

468 bạn nhé