TUYỂN TẬP TRẮC NGHIỆM TOÁN 12 THPT QG

Hien nay, co kha nhieu nguoi quan tam den no boi no thuc su rat don gian, quyen ru va ban khong can phai "hoc vet" nhieu dinh ly de co the giai duoc chung.. Khi hoc bat dang thuc, hai di

DIEN DAN TOAN HOC VIET NAM

1 Võ Quốc Bá Cần (nothing)

2 Ngo Dire Loc (Honey_ suck)

3 Trần Quốc Anh (nhocnhoc)

4 Seasky

5 Materazzi

Tran Quoc Luat’s Inequalities

Vo Quoc Ba Can - Pham Thi Hang

February 25, 2009

i

Copyright ©) 2008 by Vo Quoc Ba Can

Preface

"Life is good for only two things, discovering mathematics and teaching mathematics.”

S Poisson

Bat dang thuc la mot trong linh vuc hay va kho Hien nay, co kha nhieu nguoi quan tam den no boi no thuc

su rat don gian, quyen ru va ban khong can phai "hoc vet" nhieu dinh ly de co the giai duoc chung Khi hoc bat dang thuc, hai dieu cuon hut chung ta nhat chinh la sang tao va giai bat dang thuc Nham muc dich kich thich su sang tao cua hoc sinh sinh vien nuoc nha, dien dan mathsvn da co mot so topic sang tao bat dang thuc danh rieng cho cac ca nhan tren dien dan Tuy nhien, cac topic do con roi rac nen ta can mot su tong hop lai thong nhat hon de cho ban doc tien theo doi, do la li do ra doi cua quyen sach nay Quyen sach duoc trinh bay trong phan chinh bang tieng Anh voi muc dich giup chung ta ren luyen them ngoai ngu va co the gioi thieu no den cac ban trong va ngoai nuoc Mac du da co gang bien soan nhung sai sot la dieu khong the tranh kho1, rat mong nhan duoc su gop y cua ban doc gan xa Moi su dong gop y kien xin duoc gui ve tac gia theo: babylearnmath@yahoo.com Xin chan than cam on!

Quyen sach nay duoc thuc hien vi much dich giao duc, moi viec mua ban trao doi thuong mai tren quyen sach nay deu bi cam neu nhu khong co su cho phep cua tac gia

Vo Quoc Ba Can

1H

IV Preface

2 Let a,b,c be nonnegative real numbers such that a? + 5? + c? + abe = 4 Prove that the following inequality holds

a+h+ec > a?b? + bˆc? + cđ

3 Show that for any positive real numbers a,b,c, we have

a+bh+c+b6abe > Vabc(a+b+e)

4 Let a,b,c be nonnegative real numbers with sum 1 Determine the maximum and minimum values of

P(a,b,c) =(1+ab)? + (1+ hc)? + (1+ ca)?

5 Let a,b,c be nonnegative real numbers with sum | Determine the maximum and minimum values of

P(a,b,c) = (1 —4ab)* + (1 — 4c)? + (1 —4ca)’

6 Let a,b,c be positive real numbers Prove that

b+c eta atb ? 1 1 1

( + 5 + ¬ ) > 4(ab-+-be +ea) (+5).

Show that if a, b,c are positive real numbers, then

a " b " c "3, Pb+ bet Ca—3abe

a+b b+c c+a/j — 4 (a+b)(b+c)(c+a) `

Let a,b,c be positive real numbers Prove that

(+b \(b +e) +a’) e+e +2 \7

§a2b7c2 — \ab+bc+caj/ `

Let a,b,c be positive real numbers Prove the inequality

(b+e)" (c-+a)" (a+)

a(b+c+2a) b(c+a+2b) c(a+b+2c) —

Let a,b,c be positive real numbers Prove the inequality

a(b+c+2a) b(c+a+2b) c(a+b+2c) — \b+c+2a c+a+2b a+†b+2cj,

Let a,b,c be positive real numbers Prove that

(b+c) (eta) (a+b)? sof 4 b "_

a(b+c+2a) b(c+a+2b) c(a+b+2c) — \b+c ct+a a+bj

Let a,b,c be positive real numbers Prove that

ab+het+ea > (b+c—a)(e+a—b)(at+b—c)(a +h +c°).

a+b`+ể>——+——+_——+ b+c cta atb 2

Given nonnegative real numbers a,b,c such that ab+ bc+ca+abc = 4 Prove that

a +b? +c? +4+2(a+b+c)+3abe > 4(ab+be + ca)

Let a,b,c be real numbers with min {a,b,c} > 7 and ab + bc + ca = 3 Prove that

atbh+c+9abe > 12

Let a,b,c be positive real numbers such that a*b* + b*c? +¢7a? = 1 Prove that

(a? +b? +0°)? +abey/(a2+b2+02)3 > 4

Show that if a,b,c are positive real numbers, the following inequality holds

(a+b+ 6) (ab + be + ca)Ÿ + (ab + be + ca)3 > 4abe(a-+b+ e)Ÿ

Let a,b,c be real numbers from the interval [3,4] Prove that

b b

(atbt+e) (PO 4 ~4")\ 53/240? +2), C a b

Given ABC is a triangle Prove that

8 cos” A cos’ Bcos”C + cos2A cos2Bcos2C > 0

Let a, b,c be positive real numbers such that a+ 6+ =3 and ab+ be+ca > 2 max {ab, bc, ca} Prove that

a+h+ec > a?b? + bˆc? + cđ

Problems

Chapter 2

Solutions

"Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis?”

P Halmos, / Want to be a Mathematician

Problem 2.1 Given a triangle ABC with the perimeter is 2p Prove that the following inequality holds

p-a p- "5+ pc ¬ at ‘Vp ~c Solution Setting x = p—a,y= p—bandz= p—c, thna=y+z,b =z+x andc=x+y The original inequality becomes

+Z Z4+X xX+ +Z Z+X x+

yee poy s/o peg op x y Zz x y fog Zz

By AM-GM Inequality, we have

We have to prove

5

6 Solutions

or equivalently,

y+Z + z†# + x1 > 6,

x y Zz which 1s obviously true by AM-GM Inequality

Equality holds if and only ifa=b=c L] Problem 2.2 Let a,b,c be nonnegative real numbers such that a* + b? +c? + abe = 4 Prove that the fol- lowing inequality holds

Leer ary = Lore ee

which 1s obviously true because

>przeinet SJ6r2eler3 = VEDjer3” Cục cyc Cục

and

Ler ary Leper

Our proof is completed Equality holds if and only if a=b=c=1 ora=b= vV2,c =0 and its cyclic

which leads us to prove the sharper inequality

6( +b> +03 +6abc) > (at+b+cP4+9VeR (a+b +c),

or

S(a +b> +07) — 3) ab(a+b) + 30abe > 9Ÿ222c2(a+ b+ e),

cyc

From Schur’s Inequality for third degree, we have

3(a +b? +03) — 33 _ab(a+b) + 9abe > 0,

Cục

and we deduce our inequality to

2(a° +b? +0?) +2labe > WVeR(atb+e),

Again, the Schur’s Inequality for third degree shows that

4(a` + bŸŠ+c})+ 15abe > (a+b+e)Ÿ,

and with this inequality, we finally come up with

(a+b+c)? — Sabe +42abe > 18V.a2b2c2 (a+b +c), (at+b+c)> +27abe > 1842b2c2(a-+b+ c),

By AM-GM Inequality, we have that

2(a+b+c)> + 54abe (at+b+c)> + [(at+b+c)* +27abe + 27abc|

(a+b+e))+27a2»2c2(a+b¬+c) 9Ÿ/222c2(a+b-e)-+27Ÿa2b2c2(a+b+ e) 36VERC(a+b+e)

It shows that

(at+b+c)> +27abe > 1842b2c2(a-+b+ c),

which completes our proof Equality holds if and only ifa=b=c L] Solution 2 (by Seasky) Since the inequality being homogeneous, we can suppose without loss of generality that abc = 1 In this case, the inequality can be rewitten in the form

8 Solutions

So, the above statement holds and all we have to do is to prove that P(t,t,c) > 0, which is equivalent to each

of the following inequalities

P(a,b,c) =(1+ab)? + (1+ bc)? + (1+ ca)’

Solution It is clear that minP = 3 with equality attains when a = 1,b = c = 0 and its cyclic permutions Now, let us find max P We claim that max P = vi attains when đ — 5 = c=— 3 OF

100

(1 +ab)? + (1 +bce)? + (1+ ca)? < Te

19 aˆbŸ + bˆcˆ + c°a? + 2(ab + be + ca) < 57°

(ab + bc + ca)“ + 2(ab + be + ca) — 2abc < 7

2 46

(ab + be + ca-+ L)“ — 2abec < 7

According to AM-GM Inequality, we have

A(ab + bc + ca) — 6abc < 9°

which is obviously true by Schur’s Inequality for third degree,

1 10 A(ab + bc + ca) — 6abe < (1+ 9abc) — 6abe = 1+ 3abe < 1+3: WO

With the above solution, we have the conclusion for the requirement is minP = 3 and max P = = L]

Problem 2.5 Let a,b,c be nonnegative real numbers with sum | Determine the maximum and minimum values of

P(a,b,c) = (1 —4ab)* + (1 —4be)? + (1 —4ca)?

Solution (by Honey_suck) Notice that

with equality holds when a=b=c= 3 And we conclude that min P = 3

Problem 2.6 Let a,b,c be positive real numbers Prove that

b+ + +b ( ¬ cle a 5 + a ) >4(eð+be+ea) (TS +555) : 1 11

Assuming that a > b > c, then using AM-GM Inequality, we have that

A4(ab + be + ca)(aˆbŸ + bˆc? + c?a”) = 16(a+b)?(ab+ be+ca)(a?b* + bc? + ca’)

A(a+b) [(a+ 6)? (ab + be + ca) +4(a°b? + bc? +.c7a’)| °

It suffices to show that

(a+b)? (ab+ be+ca)+4(ab* +b’?c? +c’) 2ab(a+b) + 2be(b +) + 2ca(c +a) > ab )

10 Solutions

2 J,2 2 2 2 2

2ab(a+ b) +2c*(a+b) + 2c(a’ +b”) > (a-+6)(ab-+be-+ ca) +A Fe Fee) a

2 p2 2 (22 L p2 ab(a+b) +2c?(a+b)+c(a—b)* > 4a Aca +6")

For all real numbers m,n, p,x, y,z, we have the following interesing identity of Lagrange

(x+y +27) (me? +? + p) — (mx+ ny + pz)” = (my — nx)’ + (nz — py)? + (px— mz)’

Now, applying this identity with x = Vab,y = Vbc,z = \/ca,m = (a+ b)Vab,n = (b+c)vbc, and p = (c+a),/ca, we obtain

which 1s obviously true because

(a+b)(a+c) = yetoyvare) (a+ b)(a+c)

Problem 2.8 Given a triangle with sides a,b,c satisfying a” + bˆ -} cˆ = 3 Show that

a+b b+e c+a >6

Va+b-c Vb+c-a Vvc+a-b_ `

Solution Firstly, to prove the original inequality, we will show that!

4a(b+e—a) 4b (cta—b) 4c(at+b—c)

ab+be+ca>

4a’ (b+ce—a)

|e ere Cục >a? +b? +? —ab—be—ca,

'We may prove this statement easily by using tangent line technique, the readers can try it! In here, we present a nonstandard proof

for it, this proof seems to be complicated but it 1s nice about its idea

(b4+e? | (chap ~ (a+b420) ~ 2’

which yieds that

From (1), (2) and (3), we obtain

a’(2a—b—c) _ b(2b-c-a)? —(a’ —ab+b’) > (a+b)*(a+b— 2e)?

Using this inequality, we have to prove

(a+Ð)7(a+b—2e) 1 2 €(a+b—2c) _ 5

ae ae 3a+p+2o5 at) +—~ Gap SC (+0), (ạibp21 Ác 5 2 b

2 _ 222 ¿2 —2¿\2 (a+b)“(a+b—2c)“ _ đ(a+b—2c) > Tq„+p—2e)

(a+b) "- > I 2(a+b+2c)? (a+b)? ~ 4’

which can be easily checked Thus, the above statement is proved

Now, turning back to our problem, using Holder Inequality, we have

2

(cA) [eshte]: (m).

Moreover, by AM-GM Inequality, we have

Ya = 3 (Le) (Le) (De) 3

+ + > 3

VWb+c-a vwc+a-b va+b-c -

Solution (by Materazzi) Applying Holder Inequality, we obtain

(a+b+c) > 9[2(ab + be+ ca) — 3)

Setting p =a+b-+e, then it /3 < p <3 and 2(ab+bce+ca) = p* —3 Thus, we can rewrite the above inequality as

? > op a 6), p`—9p?+54 >0, (3 — p)(18 + 6p — p*) > 0,

which is obviously true Equality holds if and only ifa=b=c= 1 L]

14 Solutions

Problem 2.10 Show that ifa,b,c are positive real numbers, then

a + b + c S14 2abe a+b b+c c+ta™~ (a+b)(b+c)(c+a)

Solution The given inequality is equivalent to

2abe 2abe T2 at B+o(e+a) Ì a+901+o(+a)

Pu aha 2) =

Using the known inequality?

2abe Yop _ (a+b)(b+e)(e+a)'

cyc

we can deduce it to

Gab) are? (a+b)(b+c)(c+a) (a+b)(b+c)(c+a)’

Cục

abh+b’ce+ca+abe > \/2abc(a+b)(b+e)(c +a)

Now, we assume that c = min {a,b,c}, applying AM-GM Inequality, we have

ad °b+b°c+ca+abec = alab+c*)+hc(a+b)

+c)(b+

— a(a + ©)(b+‡ ©) x 9) + belatb) + a(a+c)(b+c)

? The proof will be left to the readers

15 Hence, the given inequality can be rewritten as

Tụ 4abc ` ab + be + ca

4 (a+bð)(b+c)(c+a) —(a+Ð)} (b+c) (c+a)?

a—b\' + b—c\? +(——} >2-— _ c—a\ l6abe

l6abe _ (a+b)(b+c)(c+a) 2|[(a+ b)(b+c)(c +a) — 8abc|

(a+b)(b+c)(c+a)

2l + c)(a— b)(a— e) + (c+a)(b— e)(b —a) + (a+ b)(e— a)(e— 4)

(a+ b)(b+c)(c+a) _ 2 | aes (b—c)(b—a) oe):

(at+b)(at+c) (b+c)(b+a) (c+a)(c+bÐ)

ath b+e cta

which is obviously true Equality holds if and only if@ = 6 or b=c orc=a L]

Problem 2.12 Leta,b,c be positive real numbers Prove that

(a +b*)(b? Fe*)(c* +a") 5 C4+RP4+e2\"

16 Solutions

(4a? + 1)(a+x) > 16ax(aˆ + 1— 2x), 32ax? — (16a` — 4a” + 16a— 1)x+a(4aˆ +1) >0, 2a(4x— 1)? + 2a(§x— 1) — (16a` — 4a” + 16a— 1)x+-a(4a” +1) >0,

2a(4x — 1)? + (1+ 4a’ — 16a*)x + a(4a? — 1) > 0, 2a(4x — 1)? — (2a— 1)(8a” + 2a + 1)x+a(4a? — 1) > 0,

which 1s true because

—4(2a —1)(8a* +2a4+ 1)x+4a(4a?—1) > 4a(4a* —1)— (2a—1)(8a* +2a+4+1)

(WP 4VIVP F224) =P FV 4ZICV FVLP 47 P)— Ve

Thus, we can rewrite the above inequality as

(eV tyr +27x°) [x+ty +2)“ +y +2”) 822 ty 22 +z+”)| >x43//2ˆ(x+ey +2)”

Now, we see that

(x+ty+2) “+ /+z7) —8(x 9” +y2Z +z7x”)

= yx" + 2) xy(x7 +y")+ 2xyz) x — 63x” y

= yx (x—y)(x—z) +3) xy(x — yy? +xyz) x > xyz) x

It suffices to prove that

xyo(xry $y Ze +2x°)(x+by+z) > xy (xtytz/’,

Py +y22 4+27x? > xyz(xty+z),

which 1s obviously true by AM-GM Inequality L] Solution 3 Similar to solution 2, we need to prove that

(FYI +22 42°) (ety tz) > 80P¥ tye +2")

By AM-GM Inequality, we have

(x+y+z) = x7 4y? +27 42xy+2yz+ 22x

4x7? 4y7z? 4zˆx?

IV x+y $2 +

17 Hence, it suffices to prove that

Solution 4 ” Again, we wIll give the solution to the inequality

(FYI +22 42°) (ety tz) > 80P¥ tye +2")

Assuming that x > y > z, then by Cauchy Schwarz Inequality, we have

It suffices to prove that

syle ty) 202-432) + abe tye t2)(y-+2) + Oa 4 SER WVF 5 42? Yr? 422),

8 8(x +y) 3(x —y)? 21x? — 10xy+21y”) 2?

xy form 2 1) đợi + xyz rty+ vane | — = + ) +23(x+y) > 0

>This proof seems to be the most complicated but the idea is very interesting.

18 Solutions

We have

xy | (x+y) 4/2(x?2 +32) — Aap] > xz | (xt y)4/2(x2 +32) —4xy] - òồ

Hence, it suffices to prove that

x—y)? x? — 10x 2) 2 f(z) =x l2) 2(x+y”) Ay + xy ety Than | — (21 oe ) +Z7(x+y) >0 Also, we have

ƒ(z) - fb) = § (y—z) (21x? + 13° — 18xy— 8xz — 8yz) > 0

2x(x? +y? +4xy) y(10x + 13y) ` 2x(x? +7 + 4xy) _ #(10x+ l3y)

(x+z)v2@?+y?)+4o 8x1y) 2 13)+42w 8x+y)

_— 8xÌ+22xy-—I5xy?— I3yỶ >0

Solution 5 (by Gabriel Dospinescu) We rewrite the inequality in the form

xz(x+y+z) >3 xÌ(y+z—*), cyc

It suffices to show that

which is just Schur’s Inequality for fourth degree