Gii mt s dng ton lin quan n bt ng thc hm

bng phng php chuyn qua gii hn

Trnh o Chin

Trng Cao ng S phm Gia Lai

Phng php chuyn qua gii hn dy s i khi kh hu hiu trong vic gii mt

s dng ton lin quan n bt ng thc hm. Bi vit cp n phng php ny

thng qua mt s bi ton minh ha.

1. Mt s dng ton gii bt phng trnh hm.

Bi ton 1.1. Tm tt c cc hm f : R R+ tha mn cc iu kin sau

f (x) 1 + x, x, y R; (1)

f (x+ y) f (x) f (y) , x, y R. (2)

Gii.

Trc ht, lu rng f (x) > 0, x R. Trong (1), cho x = 0 ta c f (0) 1.Trong (2), cho x = y = 0 ta c f (0) f 2 (0), suy ra f (0) 1. Do f (0) = 1.

iu kin (2) suy ra rng f (x1 + x2 + ...+ xn) f (x1) f (x2) ...f (xn), vi mixi R, 1 i n. Do f (x) = f

(xn+ ...+ x

n

) fn

(xn

), vi mi x R v n N.

Kt hp vi iu kin (2), ta c f (x) fn(xn

)(1 + x

n

)n. T bt ng thc ny, cho

n , ta c f (x) ex. Hn na, ta c 1 = f (0) = f (x+ (x)) f (x) f (x) exex = 1.

Do f (x) = ex. Th li ta thy hm ny tha mn iu kin bi ton.

Bi ton 1.2. Chng minh rng khng tn ti hm f : R+ R+ sao cho

f (x+ y) f (x) + y.f (f (x)) , x, y R+. (3)

Gii.

Gi s rng tn ti hm f tha mn iu kin ca bi ton.

Trong (3), cho x = 1 v thay y bi x ta c f (1 + x) f (1) + x.f (f (1)). iuny suy ra rng lim

x+f (x) = +, v do lim

x+f (f (x)) = +.

Mt khc, trong (3), cho y = 1, ta c

f (x+ 1) f (x) + f (f (x)) , x R+. (4)

iu ny suy ra rng limx+

(f (x+ 1) f (x)) = +. Do , tn ti x0 R+ saocho

f (x0 + k) f (x0 + k 1) > 2, k 1. (5)

1

By gi, chn mt gi tr xc nh n N sao cho n x0 +1. Trong (5), cho k lnlt nhn cc gi tr 1, 2, ..., n v sau cng cc bt ng thc thu c, ta c

f (x0 + n) f (x0) > 2n, n x0 + 1. (6)

Hn na, v f (x0) > 0 nn vi cch chn n x0 + 1, ta c n > x0 + 1 f (x0). Do, bi (6), ta c

f (x0 + n) > 2n+ f (x0) > x0 + n+ 1. (7)

Th th, ta c

f (f (x0 + n)) f (x0 + n+ 1) + (f (x0 + n) (x0 + n+ 1)) .f (f (x0 + n+ 1)) . (8)

V f (f (x0 + n+ 1)) > 0 nn, bi (7), ta c

f (x0 + n+ 1) + (f (x0 + n) (x0 + n+ 1)) .f (f (x0 + n+ 1)) > f (x0 + n+ 1) . (9)

Hn na, bi (4), ta c

f (x0 + n+ 1) f (x0 + n) + f (f (x0 + n)) . (10)

Ngoi ra, v f (x0 + n) > 0 nn ta c

f (x0 + n) + f (f (x0 + n)) > f (f (x0 + n)) . (11)

Cui cng, t cc bt ng thc (8), (9), (10), (11), suy ra f (f (x0 + n)) >

f (f (x0 + n)), mu thun. Ta c iu phi chng minh.

Bi ton 1.3. Tm tt c cc hm s lin tc f : [0; 1] R tha mn iu kin

f (x) 2xf(x2), x [0, 1] . (12)

Gii.

Trong (12), thay ln lt x = 0 v x = 1 ta c

f (0) 0, f (1) 0. (13)

Vi x [0;

1

2

), p dng (12) n ln, ta c

f (x) 2xf(x2) 22x3f

(x4) ... (2x)nx2nn1f

(x2

n), n N. (14)

V x [0;

1

2

)v f lin tc nn

limn+

((2x)nx2

nn1f(x2

n))= 0. (15)

2

T (14) v (15), ta c

f (x) 0, x [0;

1

2

). (16)

Mt khc, vi x (0; 1), bi (12), ta c f (x) 2

xf (x). Suy ra

f (x) f (x)

2x ...

f

x 12n

2nx1

1

2n

. (17)

M limn+

f

x 12n

2nx1

1

2n

= 0, nn bi (17), ta c

f (x) 0, x (0; 1) . (18)

T (16) v (18), ta c

f (x) = 0, x [0;

1

2

). (19)

Vi mi x [1

2; 1

), tn ti n N sao cho x2n < 1

2. Th th, bi (14) v (19), ta

c f (x) 2nx2n1f(x2

n)= 0. Do

f (x) 0,x [1

2; 1

). (20)

Bi (18) v (20), ta c

f (x) = 0, x [1

2; 1

). (21)

Bi (19) v (21), suy ra f (x) = 0, x [0; 1).Hn na, v hm f lin tc trn [0; 1] nn f (x) = 0, x [0; 1].Th li, ta thy f (x) = 0, x [0; 1] tha mn iu kin bi ton.Bi ton 1.4. Xt bt phng trnh hm

f (x+ y) f (x) g (y) + f (y) g (x) , x, y R, (22)

trong g (x) l mt hm gii ni, kh vi ti 0, g (0) = 1 v g (0) = k. Chng minh

rng f (x) 0 l hm s duy nht tha mn bt phng trnh cho, vi iu kin

limx0

f (x)

x= 0. (23)

Gii. Gi s rng f (x) l nghim ca (22), vi iu kin (23).

3

Th th, vi h > 0 nh, ta c f (x+ h) f (x) g (h)+f (h) g (x) hay f (x+ h)f (x) (g (h) 1) f (x) + f (h) g (x). Do

f (x+ h) f (x)h

g (h) g (0)h

f (x) +f (h)

hg (x) .

Mt khc, ta c f (x) = f (x+ h h) f (x+ h) g (h) + f (h) g (x+ h) hayg (h) (f (x) f (x+ h)) g (h) f (x) f (x) + f (h) g (x+ h) .

V hm g (x) kh vi ti 0 nn n lin tc ti im . Do , vi h > 0 nh, ta

c g (h) > 0. Vy, vi h > 0 nh, ta c

f (x+ h) f (x)h

(g (h) 1) f (x) + f (h) g (x+ h)g (h)

=g (h) g (0)h.g (h)

f (x) +f (h)h.g (h)

g (x+ h) .

Vy vi h > 0 nh, t cc kt qu trn, ta c

g (h) g (0)h

f (x) +f (h)

hg (x) f (x+ h) f (x)

h

g (h) g (0)h.g (h)

f (x) +f (h)h.g (h)

g (x+ h) .

Tng t, bt ng thc trn cng ng i vi chiu ngc li, vi h < 0 nh. Do

, bi iu kin (23), ta c

f (x) = limh0

f (x+ h) f (x)h

tn ti v bng g (0) f (x) = kf (x), vi x R, v g (x) l mt hm gii ni.T , vi x R, ta c(

f (x)

ekx

)=f

(x) kf (x)

ekx=kf (x) kf (x)

ekx= 0.

Do f (x) = Cekx (C l hng s). Hn na, t iu kin (2) suy ra rng C = 0.

Vy f (x) 0 l hm s duy nht tha mn bt phng trnh cho, vi iu kin(23).

2. Mt s dng ton lin quan n bt ng thc hm.

Bi ton 2.1. Gi s f : R+ R+ l hm tha mn iu kin

f (2x) > x+ f (f (x)) , x R+. (24)

Chng minh rng f (x) > x, x R+.

4

Gii. Bi (24), ta c

f (x) >x

2+ f

(f(x2

))>x

2, x R+. (25)

Gi s rng

f (x) > anx, x R+, (26)

trong an l hng s. Th th, bi (24), (25), (26), ta c

f (x) >x

2+ f

(f(x2

))>x

2+ anf

(x2

)>

1 + an2

2x.

Xt dy (an)n=1 xc nh bi

a1 =1

2, an+1 =

1 + an2

2, n > 1.

Th th an+1 an =(1 an)2

2> 0, ngha l (an)

n=1 l mt dy tng. Hn na,

d thy rng an < 1, vi mi n > 1. Suy ra dy l hi t v nu k hiu a l gii

hn ca n th a =1 + a2

2, ngha l a = 1. Do , cho n th t bt ng thc

f (x) >1 + an

2

2x, ta c f (x) > x, x R+. Ta c iu phi chng minh.

Bi ton 2.2. Gi s f : R R l hm tha mn cc iu kin

f 2 (x) 2x2f(x2

),x R; f (x) 1, x (1, 1) .

Chng minh rng f (x) x2

2, x R.

Gii.

D thy rng f (0) = 0. Do , ta ch cn chng minh bt ng thc vi x 6= 0.

t g (x) =2f (x)

x2, vi x 6= 0. Th th g2 (x) g

(x2

)v do g2

n(x) g

( x2n

), vi

x 6= 0 v n N. Ch rng, g (x) g2 (2x) 0. Do , ta c

g (x) 2ng( x2n

)= 2

n

2f( x2n

)( x2n

)2 2n

22n+1

x2,

vx

2n (1, 1). By gi cho n v s dng kt qu lim

n

n

2n= 0, ta thu c

g (x) 1. Do f (x) x2

2, x R.

Bi ton 2.3. Gi s F l tp tt c cc hm s f : R+ R+ tha mn bt ngthc

f (3x) f (f (2x)) + x, x R+. (27)

5

Tm s thc a ln nht sao cho vi mi hm f F , ta lun c

f (x) ax. (28)

Gii.

D thy rng f (x) =x

2tha mn (27), nn f (x) =

x

2 F . Thay f (x) = x

2vo

(28), ta suy ra a 12.

V f (x) > 0, x > 0, nn t (27) ta c

f (x) = f

(3x

3

) f

(f

(2x

3

))+x

3>x

3, x > 0. (29)

Ta xc nh mt dy (an) nh sau

a1 =1

3; an+1 =

2an2 + 1

3, n 1. (30)

D dng kim tra rng

0 < an 0, n 1.

Do , (an) l dy s dng, tng nghim ngt v b chn bi1

2.

Vy dy (an) hi t. Gi s limn

an = . Th th, bi (30) v (31), ta c =22 + 1

3

hay =1

2.

By gi ta cn chng minh rng, vi mi n 1, ta lun c

f (x) anx, x > 0. (32)

Tht vy, bi (29) nn (32) ng vi n = 1. Gi s (32) ng vi n = k 1, nghal

f (x) akx, x > 0. (33)

Khi , bi (27) v (33), ta c

f (x) f(f

(2x

3

))+x

3 ak.f

(2x

3

)+x

3

ak2.2x

3+x

3=

2ak2 + 1

3.x ak+1x, x > 0.

Vy (32) ng vi n = k + 1. Do , (32) ng vi mi n 1.

6

T cc iu trn suy ra rng, vi mi hm f F , ta lun c f (x) 12x.

Tm li, gi tr ca a cn tm l a =1

2.

Bi ton 2.4. Tm tt c cc hm s f : [1,) [1,) tha mn cc iu kinsau

f (x) 2 (1 + x) , x 1; (34)

xf (x+ 1) = f 2 (x) 1, x 1. (35)

Gii. Bi cc gi thit (34) v (35), ta c

f (x) =xf (x+ 1) + 1

2x (x+ 2) + 1

x, x 1.

Bng phng php quy np, ta chng minh c rng

f (x) > x1

1

2n , x 1, n 1. (38)

V1

2n 0 khi n, nn bi (38) ta c

f (x) x, x 1. (39)

By gi, bi (35) v (39), ta c

f (x) =xf (x+ 1) + 1

x (x+ 1) + 1 > x+

1

2, x 1.

Bng phng php quy np, ta chng minh c rng

f (x) > x+

(1 1

2n

), x 1, n 1. (40)

V1

2n 0 khi n, nn t (40) ta c

f (x) x+ 1, x 1. (41)

7

Bi (37) v (41), ta suy ra f (x) = x+ 1, x 1.Th li, ta thy hm s f (x) = x+ 1, x 1, tha mn iu kin bi ton.

Bi tp.

Bi 1. Gi f : R R l hm s tha mn iu kin

|f (x+ y) f (x) f (y)| 1, x, y R.

Chng minh rng tn ti hm s g : R R tha mn cc iu kin sau

i) |f (x) g (x)| 1, x R;

ii) g (x+ y) = g (x) + g (y) , x, y R.

Hng dn gii. Hm g (x) = limn+

f (2nx)

2nl hm cn tm.

Bi 2. Chng minh rng khng tn ti hm s f : R R tha mn iu kin

f (x) + f (y)

2 f

(x+ y

2

)+ |x y| , x, y R.

Hng dn gii. Gi s tn ti hm s f tha mn iu kin bi ton. Ta chng

minh c bt ng thc

f (x) + f (y)

2 f

(x+ y

2

)+ 2n |x y| , x, y R, n 0.

V 2n +, khi n +, nn dn n iu mu thun.Bi 3. Cho hm s f : R R+ tha mn iu kin

3f (x)

3f (x) 9

4f

(4x

3

) 1, x R.

Tm s thc k ln nht sao cho f (x) k, vi mi x R.Hng dn gii. Xt dy s (un) xc nh nh sau

u1 =4

9, un+1 =

4

9+

8

9

3un, n 1.

Dy s ny tng v b chn trn nn tn ti gii hn, gii hn l4

3. Suy ra k =

4

3.

Gia Lai, 8/ 3/ 2013

T..C

8

TI LIU THAM KHO[1]. Nguyn vn Mu, Bt ng thc, nh l v p dng, Nh xut bn Gio dc,

2006.

[2]. IH-Ching, On some functional inequalities, 129 - 135, Aequationes Mathemat-

icae, Vol. 9, 1973.

[3]. Titu Andresscu, Iurie Boreico, Functional equations - 17 Chapters and 199

Problems with solution, Electronic Edition, 2007.

[4]. Mt s Tp ch Ton hc v Tui tr.

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