1
1.Let a,b,c be nonnegative real numbers,no two of which are zero.Prove that
a a 2+bc +
b b 2+ca +
c c 2+ab ≥√
23
(a +b +c )2.
(Vo Quoc Ba Can,inspired from Shalex’s one)
Solution.Firstly,we will prove that
cyc
大陆泽
a
2(a 2+bc )≥
cyc
a (a +
信贷配给b )(a +
c )⇔ cyc
a 2(a 2+bc )− (a +
b )(a +
润肺止咳汤c ) ≥0
⇔
cyc
a (a −
令主b )(a −
学官c )
2(a 2+bc )+ (a +b )(a +c )
≥0
Without loss of generality,we may assume a ≥b ≥c ,then we have
a (
b +
c )(b +a )≥b (a +b )(a +c ),a 2(b 2+ca )≥b
张家界的景点2(a 2+bc )Hence
a
2(a 2+bc )+
(a +b )(a +c )
≥b
2(b 2+ca )+
(b +c )(b +a )
Thus
cyc
a (a −
b )(a −
c )
2(a 2+bc )+
(a +b )(a +c )≥(a −b ) a 2(a 2+bc )+ (a +b )(a +c )−b 2(b 2+ca )+ (b +c )(b +a )
≥0
Now,we will prove that
cyc
a (a +
b )(a +
c )≥
2
3
cyc
a
2
扶不起
⇔
cyc
a 2(a +
b )(a +
c )+2
cyc
ab ≥
49
cyc
a
4By the Cauchy Schwarz Inequality,we have
(a +b )2(a +c )(b +c )=(a 2+ab +bc +ca )(b 2+ab +bc +ca )≥(ab +ab +bc +ca )2
2
It suffices to show that
cyc a2(a+b)(a+c)+2
cyc
ab(2ab+ac+bc)≥
4
9
cyc
a
4
Due to the homogeneity,we may assume a+b+c=1.Setting q=ab+bc+ ca,r=abc,then we have
cyc a2(a+b)(a+c)=
cyc
a2(a+bc)=1−3q+4r
cyc ab(2ab+ac+bc)=2
cyc
ab
2
−2
cyc
a2bc=2q2−2r
The inequality becomes
1−3q+4r+2(2q2−2r)≥4 9
⇔1
9
(1−3q)(5−12q)≥0.
which is trivial since q≤1
3.
The inequality is proved.Equality holds if and only if a=b=c.
♥♥♥
Vo Quoc Ba Can
Student Can Tho University of Medicine and Pharmacy,Can Tho,Vietnam
梦见杀人流血
E-mail:can−
