1.Algebra Booster

Sequence and Series 1.53
169. Applying AM ≥ GM, we get
(1 - x) + (1 - y) + (1 - z)
≥ 3 (1 - x)(1 - y)(1 -z)
3
fi
3 - ( x + y + z)
≥
3
(1 - x)(1 - y)(1 -z)
3
fi
3- 1 ≥ 3 (1 - x)(1 - y)(1 - z)
3
fi
2
≥ 3 (1 - x)(1 - y)(1 -z)
3
fi
8
(1 - x)(1 - y)(1 - z)
£
27
Hence, the result.
170.
1 1 1 c + a + b 1
+ + = =
ab bc ca abc abc
(given)
We know that
Êa + b + cˆ 3
Á ≥ abc
Ë
˜
3 ¯
fi
3
Êa + b + cˆ Á ≥ abc
Ë
˜
3 ¯
fi
1
abc £
27
fi
1
abc ≥ 27
fi
Ê 1 1 1 ˆ
Á + + ≥27
Ë
˜
ab bc ca¯
Hence, the result.
171. We have
(ax + by) 2 – (a 2 + b 2 )(x 2 + y 2 )
= a 2 x 2 + b 2 y 2 + 2abxy – a 2 x 2 –a 2 y 2 – b 2 x 2 – b 2 y 2
= –(a 2 y 2 + b 2 x 2 – 2abxy)
= –(bc – ay) 2
= –ve
Thus, (ax + by) 2 < (a 2 + b 2 )(x 2 + y 2 )
fi (ax + by) 2 < 1 ◊ 1 = 1
fi (ax + by) < 1
Hence, the result.
172. Applying AM ≥ GM, we get
Ê px + qyˆ Á ≥
Ë
˜
2 ¯
pqxy
fi ( px + qy) ≥2
pqxy
Similarly, ( pq + xy) ≥2
pqxy
Thus, (px + qy)(pq + xy) ≥ 4pqxy
Hence, the result.
173. We have,
log 2
x + log 2
y ≥ 6
fi log 2
(xy) ≥ 6
fi (xy) ≥ 2 6 = 64
Applying AM ≥ GM, we have
Êx
+ yˆ Á ≥
Ë
˜
2 ¯
xy
fi
Êx
+ yˆ Á ≥
Ë
˜
2 ¯
64 = 8
fi (x + y) ≥ 16
Hence, the minimum value of (x + y) is 16.
174. Applying AM ≥ GM, we have
Ê logab
+ logba
Á
ˆ ≥ logab
logba
Ë
˜
2 ¯
Êlogab
+ logbaˆ fi Á
≥ 1
Ë
˜
2 ¯
fi log a
b + log b
a ≥ 2
Hence, the minimum value is 2.
175. We have,
2 log 10
x – log x
(.01) = 2 log 10
x – log x
(10) –2
= 2 log 10
x + 2 log x
(10)
= 2(log 10
x + log x
(10))
≥ 2.2 – 4
Hence, the least value of 2 log 10
x – log x
(.01)
is 4.
176. We know that,
2 2
2 2
- 1 + b
1
2 Á -
2˜
Á Ê a
ˆ Ê ˆ
Ë
˜
a ¯ Ë b ¯
Ê 4 1 ˆ Ê 4
a b
1 ˆ
= Á + + + -4
Ë 4˜ Á 4˜
a ¯ Ë b ¯
Ê
2 2 2 2 2 2
( a ) + ( b ) ˆ Êa + b ˆ
Also, Á ≥
Ë
˜ Á ˜
2 ¯ Ë 2 ¯
fi
fi
Also,
fi
Ê
2 2 2 2
( a ) + ( b ) ˆ 1
Á
≥
Ë
˜
2 ¯ 4
4 4 1
( a + b ) ≥
2
2 -2 2 -2 2 2
( a ) + ( b ) Êa + b ˆ
≥ Á
2 Ë
˜
2 ¯
2 -2 2 -2
( a ) + ( b )
2
≥ 4
2
-2
fi
Ê 1 1 ˆ
Á + ≥8
Ë 4 4˜
a b ¯
4 4
Thus,
Ê 1 1 ˆ
Áa
+ + b + -4 ≥ 1 + 8-4
Ë 4 4 ˜
a b ¯ 2
fi
2 2
2 2
b
2 2
Ê 1 ˆ Ê
a
1 ˆ
Á - + - ≥
9
Ë
˜
a ¯
Á
Ë
˜
b ¯ 2
Hence, the result
(given)