duoc : (p'q - pq)v2qu = (u'v-uv)q2pv
<+ (p'v + pv)q2vu = (q'u + qu)vzpq
Theo (4) th\qu.qv - pv.qv + 0 n€n suy ra p't' + pv' - cl'u + c1u' hay
P'(x.)V(x,,) + P(x")V'(x") =
= Q'(x.,)U(x,) + Q@)U'(x") (6)
OUriu tiu
C6c ban doat gi6i c0a Cupc thi Vui hti 2001 vit
CuQc thi gidi Todn vd VQt Li tr6n THTT ndm hqc
2OOO-2OO{ hay ti6p tuc gfri dia chi m6i c0a minn vd
,ou.o,.n.
,r*
r Ndu pu = 0 thi ti (2) c6 p = u = O,thay vho
(5) duoc p'qv2 = r.ivqz > p'v = u'q. TU didu ndy
cing vdi P = Lt = 0 cf,ng suy ra (6).
T5m lai, tir (4), (6) vd bd ad ta kdt luAn rang
phuong trinh P(x)V(x) - Q@)U(x) = 0 c6
nghiOmbOi x=xo.
DAo lai ndu x - xn vdi Q@)V(x.) + 0 lI
nghiOm bQi cira phuong trinh P(x)V(x)
Q(x)U(x) = 0 thi tit (4) vh (6) bidn ddi ngu-oc lai
ta se suy ra (2) vi (3). Theo (l) d0 thi hai hdm s6
tidp xric vdi nhau tai didm c6 hodnh d0 x =.xn. Chring tOi cho r6ng trong srich gi6o khoa nen trinh bdy dinh li n6i rr€n (cd thd khong chfng minh holc chfng minh v6i chfi nho) dd
ldm co s6 li thuydt cho phuong ph6p nghiOm
bOi. Tt d6 cho ph6p hoc sinh du-o. c str dung
phuong ph6p niy khi giAi c6c bli toi{n vd su tidp
xfc ctra dd thi hai him phAn thrlc hftu ti cflng
nhu bdi toiin vd tidp tuydn vdi dd rhi hlm phAn
thrlc hfru ti.
(2)
rrENG ANrr ouA cAc BAr ToAil
BAI Sd 52
Problem. Let x1 3 x2 A ... 3 xn and y1 t Jt a
...Sln be two non-decreasing sequence ofreal
numbers. A permutation of yt,...,Jn is an
arrangement of the terms of this sequence in a
particular order. For an arbitrary permutation !i1 , !i2..., linof y1,!2,...,y,r prove that x1 y,, +
xZ!,2+ ... * Jn !i,, S xtlt + xZJZ+ ... + xilyfi
and that equality holds if and only if x1 = x, -
... = Xn Of )t - J2= ... = in.
Solution.
Put S = xtlil+ xzliz+ ... + *rlitr.
Suppose that y,, = li1. In the sum S we
interchange !,,,und )iy . This transform S into S' =Jt1 !.i1* x2!ir+ ... * xjli,, + ... + xnyr.
We have S' - S = xn(!, - !i,,) + xj(yi, -yn) = = (xr- x)(ln - )i,, ) > 0
because xr2 xi and y,, ) )in . Thus, S' > S. Next
find in S' the summand containing Jn-t.
Suppose that y,,-1 - )t1 . In the sum S' we
interchange !ir_, and 1l;1 . This transform S'
into a new sum S". Similarly, it is easy to verify
that S" > S'. This process can be continued until
we obtain the sum xtlt * xzJz * ... I xn!, which is larger than or equal to all previous
sums S, S', S",... So we have proved thatS<xtyr + xzyz+ ... + xtJn. S<xtyr + xzyz+ ... + xtJn.
It is clear that equality holds if xy = x2 = ... =XnOf lt=JZ=...=!n.
Tir mdi:
decreasing = giam
real = thuc (tinh tt)
permutation = ph6p hodn vi term = sd h+ng
particular = d4c biQt, riCng (tinh tt) summand = hans ttl
interchange = ddi ;ho, thay rhe nhau (dOng tir)
transform = bi€n ddi (dgng tt)
similarly = mQt giich tuong tu,
cfrng nhu vay (ph5 tU) verify = x6c minh, x6c nh1n (d.ng tt)
process = qu6 trinh
continue = ti€p tqc (dOng tt)
previous = trudc (tinh tit)
NGO VIET TRUNG