1.Algebra Booster

1.58 Algebra Booster
Thus, (1 + x)(1 + y) < 2(xy + 1)
Similarly, (1 + z)(1 + w) < 2(zw + 1)
Therefore,
(1 + x)(1 + y)(1 + z)(1 + w) < 4(xy + 1)(zw + 1)
208. It can be easily proved that
(1 + x)(1 + y)(1 + z)(1 + w) < 4(xy + 1)(zw + 1)
Obviously, (1 + xy)(1 + zw) < 2(xyzw + 1)
Hence, (1 + x)(1 + y)(1 + z)(1 + w) < 8(xyzw + 1)
209. As we know that AM ≥ GM
4 4
or
a + b 4 4
≥ ab
2
fi (a 4 + b 4 ) ≥ 2a 2 b 2
fi (a 4 + b 4 + c 2 ) ≥ 2a 2 b 2 + c 2
fi
4 4 2 2 2 2 2 2 2
( a + b + c) ≥ 2ab + c ≥2 2abc
4 4 2
fi ( a + b + c ) ≥2 2abc
Hence, the result.
210. Since,
1 Êab bcˆ
ab bc
Á + ≥ ¥ = b
2 Ë
˜
c a ¯ c a
1 Êbc caˆ
bc ca
Á + ≥ ¥ = c
2 Ë
˜
a b ¯ a b
1 Êca abˆ
ca ab
Á + ≥ ¥ = a
2 Ë
˜
b c ¯ b c
Adding relations (i), (ii) and (iii), we get
ab bc ca
+ + ≥ a + b+
c
c a b
Hence, the result.
211. By the mth power theorem,
fi
Ê
2 2
b + c ˆ Êb+
cˆ
Á ≥
Ë
˜
2 ¯
Á
Ë
˜
2 ¯
Ê
2 2
b + c ˆ Êb+
cˆ
Á ≥ Á ˜
Ë b + c
˜
¯ Ë 2 ¯
Ê
2 2
c + a ˆ Êc + aˆ
similarly, Á ≥ Á ˜
Ë c + a
˜
¯ Ë 2 ¯
2
…(i)
…(ii)
…(iii)
…(i)
…(ii)
Ê
2 2
a + b ˆ Êa + bˆ
and Á ≥ Á ˜
a b ˜ …(iii)
Ë + ¯ Ë 2 ¯
Adding relations (i), (ii) and (iii), we get
Ê
2 2 2 2 2 2
b + c c + a a + b ˆ
Á + +
Ë b + c c + a a + b
˜ ≥ (a + b + c)
¯
Hence, the result.
212. By the mth power theorem, we get
fi
-2 -2 -2 2 2 2
-1
Ê ˆ Ê ˆ
x + y + z x + y + z
Á ≥
Ë
˜ Á ˜
3 ¯ Ë 3 ¯
Ê
-2 -2 -2
x + y + z ˆ Ê 3 ˆ
Á ≥
Ë
˜ 2 2 2
3 ¯ Á
Ëx + y + z ˜
¯
fi
- - - Ê ˆ
x + y + z ≥ Á = 9
Ë
˜
1¯
2 2 2 9
1 1 1
+ + ≥9
x y z
fi
2 2 2
213. a –5 + a –4 + 3a –3 + 1 + a 8 + a 10
fi
= a –5 + a –4 + a –3 + a –3 + a –3 + 1 + a 8 + a 10
Ê
-5 -4 -3 -3 -3 8 10
a + a + a + a + a + 1 + a + a ˆ
= Á
Ë
˜
9
¯
9 -5 -4 -3 -3 -3 8 10
≥ a ◊a ◊a ◊a ◊a ◊a ◊ a = 1
Êa + a + a + a + a + 1 + a + a
Á
Ë
9
-5 -4 -3 -3 -3 8 10
fi (a –5 + a –4 + a –3 + a –3 + a –3 + 1 + a 8 + a 10 ) > 1
Hence, the minimum value is 1.
LEVEL III
Ê
1. Let 1 ˆÊ
t 1 ˆ
r = Ár + r +
Ë
˜Á 2 ˜
w ¯Ë w ¯
Ê 2 Ê 1 1 ˆ 1 ˆ
= Ár
+ + r +
Ë
Á
Ë 2˜
3
w w ¯ ˜
w ¯
Ê 2 2 1 ˆ
= Ár
+ ( w + w)
r +
Ë
3 ˜
w ¯
Now,
= (r 2 – r + 1)
S
n
n= Âtr
r = 1
n
2
Â
= ( r - r + 1)
r = 1
n
2
n n
  Â
= r - r + 1
r= 1 r= 1 r=
1
nn ( + 1)(2n+ 1) nn ( + 1)
= - + n
6 2
Ê( n+ 1)(2n+ 1) ( n+
1) ˆ
= nÁ
- + 1
Ë
˜
6 2 ¯
1 ( (2
2 3 1 3 3 6))
= n n + n+ - n - +
6
Ê
2
nn ( + 2) ˆ
= Á
Ë
˜
3 ¯
ˆ
˜ > 1
¯
Hence, the result.
2. Let t r
= (r – 1)(r – w)(r – w 2 )
= r 3 – (1 + w + w 2 )r 2 + (1 ◊ w + w ◊ w 2 + w 2 ◊ 1)r
–(1 ◊ w ◊ w 2 )
= r 3 – 0 ◊ r 2 + 0 ◊ r – 1
= r 3 – 1