: (BT, BX ) (CY, CX) (mod zr)
thoa mdn diiu ki€n a2 +b2 +c2 +abc=4.
Chwng minh rdng a2 + h: + c2 -3abt'< 2(a + b + c1 . Ta thAy 4 = a2 + b2 + c2 + abc = a2 +(b +-c)z 2 *a(b+c)z *( * *o_ (b+c)'z)* o( *_@+c)'\ 4 \ 2 )'-l* 4 ) , . (2+a)(b+c\2 . (b-c)z a(b-c)2 424 . . (2 + a)(b + c)2 (2 - a)(b - c\2 -q T-. 44 vi 0 < a < 2n€n 4 ) a2 +(2+ a)(b+ c)2 4 = 4 _ oz, (2 + a)(b + c)2 = 2 _ o, (b + c)' 44 >b+c<2J2-a.
Tucrng t.u c+a12J2-b;a+b<2J2-c .
CQng theo ba BDT tr6n ta c6
a + b +, < Jz-i + Jz-t + Ji- r .
Vpy ta thu duo. c bdi to6n
ft nei tuin 4. Cho cac t6 khang dm a, b, c
th6a mdn diiu ki€n az + bz + c2 + abc = 4.ChLrng minh ring ChLrng minh ring
a + b +, < Ji* o n Jz- o * Jz-, .
Ta th6y n6u ci ba sti ducrn1 o, b, c ctng l6n hon (ho[c cing nh6 hcrn) 1 thi s€ mdu thuSn voi gi6 thi€t a2 + b2 + c2 + abc = 4 . N6n ta c6
th6 gin sir a) l, b>tr,c<1 ho[c a 7-1, b /-1,c)1. Tu d6 suy ra (a - lxb - 1) > 0 c)1. Tu d6 suy ra (a - lxb - 1) > 0
€ a+b-ab<1 (1)M{t kh5c theo B}ri to6n 3 ta c6 2-c) ab M{t kh5c theo B}ri to6n 3 ta c6 2-c) ab
suy ra 0 < c ( 2- ab (2). Tt (1) ve (2) suy ra
(a + b - ab)c < c 3 2 - ab > ab + bc + ca - abc < 2.
Hon nta c !1, n€n abc 1ab, suy ra
0 < ac + bc + ab - abc . Vpy ta thu dugc bdi to6n
fi nai tofn 5. Cho cdc ti| kharg dm a, b, c
thoa mdn diiu ki€n a2 +b2 +c2 +abc=4.
Chimg minh rdng 0<ab+bc+ca-abc<2.
(usA Mo - 2001).
Theo Cfch 3 thi ta c6 th6 d{t a : 2cosA,b=2cosB, c : 2cosC trong d6 A, B, C ld s6 b=2cosB, c : 2cosC trong d6 A, B, C ld s6
ilo cilc g6c ctra tam giitc ABC L,h6ng tu. Vay thi ta c6 th6 luqngFi6chodmQt sO bei to6n da tim dugc 0 tr€n d0 tao ra nhfrng bdi to6n bAt ding thric trong tam gi6c.
ff nei lrrhn 6. Cho tam giac ABC kh6ng tit.
, ,;
Chmtg ntinh cac bat dang thtrc sau
1) coslcosBcosC<];
8
2) cos AcasB+ cos,BcosC +.osC"osl <
];
1
3) cos2,4 +cos2 B+cos2 C > l;
4
4) cosl + cos.B + cosC+ coslcosB + cosBcosC +cosCcos A-2cosAcosBcos C 32;