Câu 5,
Ap dung BDT Cosi
[tex]\dfrac{a^2}{b-1}+4(b-1)\geq 4a[/tex]
[tex]\dfrac{b^2}{a-1}+4(a-1)\geq 4b[/tex]
Suy ra
[tex]P+4(a+b)-8\geq 4(a+b)[/tex]
[tex]P\geq 8[/tex]
MinP=8
Dau = khi a=b=2.
Giai:
Cau 4:
a, Ta co:
Goc EFB=EHB=90
Suy ra: BEFH noi tiep.
b, Tam giác ABE có I là trực tâm
Tam giác BIC dong dang BEF (g.g)
Suy ra: BI.BF=BC.BE
c,
Ta có:
$EH=BH=\dfrac{3R}{2}$
[tex]S_{EAB}=\dfrac{EH.AB}{2}=\dfrac{\dfrac{3R}{2}.2R}{2}=\dfrac{3R^2}{2}[/tex]