A small inequality problem
Yue Kwok Choy
This is a small inequality sent to me by Bowen Kwan to prove:
3
√(a1 + b1 )(a2 + b2 )(a3 + b3 ) ≥ 3√a1 a2 a3 + 3√b1 b2 b3 where ai , bi ≥ 0
…. (1)
My last teaching day with
7E class. I acted as a
substitute teacher for
this vivacious and happy class
after my retirement in 2009.
(Photo taken in 2011)
My first attempt is to use brute force to cube the inequality:
3
√(a1 + b1 )(a2 + b2 )(a3 + b3 ) ≥ 3√a1 a2 a3 + 3√b1 b2 b3
3
⇔ (a1 + b1 )(a2 + b2 )(a3 + b3 ) ≥ ( 3√a1 a2 a3 + √b1 b2 b3 )
3
⇔ (a1 a2 b3 + a2 a3 b1 + a3 a1 b2 ) + (a1 b2 b3 + a2 b3 b1 + a3 b1 b2 )
2
≥ 3( 3√a1 a2 a3 ) ( 3√b1 b2 b3 ) + 3( 3√a1 a2 a3 )( 3√b1 b2 b3 )
a1 a2 b3 +a2 a3 b1 +a3 a1 b2
⇔(
3
a1 b2 b3 +a2 b3 b1 +a3 b1 b2
)+(
3
2
)
3
3
≥ √(a1 a2 b3 )(a2 a3 b1 )(a3 a1 b2 ) + √(a1 b2 b3 )(a2 b3 b1 )(a3 b1 b2 )
which is obviously true by using A.M.  G.M. twice.
Since Bowen hinted that whether this inequality can be proved by CBS inequality, so I began my
second attempt. Using the CBS inequality for two sets of variables case looked easy:
(x1 2 + y1 2 )(x2 2 + y2 2 ) ≥ (x1 x2 + y1 y2 )2
Putting
x1 2 = a1 , y1 2 = b1 , x2 2 = a2 , y2 2 = b2 ,
(a1 + b1 )(a2 + b2 ) ≥ (√a1 a2 + √b1 b2 )
or
Although
we get:
2
√(a1 + b1 )(a2 + b2 ) ≥ √a1 a2 + √b1 b2
…. (2)
(2) is not we wanted, but it is a good step forwards to have my proof.
P.1
Applying the C.B.S. inequality directly is not fruitful for proving (1), the reader may try.
It is lucky that we can extend to 4 sets of variables case:
4
√(a1 + b1 )(a2 + b2 )(a3 + b3 )(a4 + b4 ) = √√(a1 + b1 )(a2 + b2 )√(a3 + b3 )(a4 + b4 )
≥ √(√a1 a2 + √b1 b2 )(√a3 a4 + √b3 b4 )
≥ √√a1 a2 √a3 a4+√√b1 b2 √b3 b4
= 4√a1 a2 a3 a4 + 4√b1 b2 b3 b4
…. (3)
To get back to (1), we need to put away a4 , b4 .
3
a4 = 3√a1 a2 a3 , b4 = √b1 b2 b3
We let
and substitute in (3) .
4
√(a1 + b1 )(a2 + b2 )(a3 + b3 )(a4 + b4 ) ≥ 4√a1 a2 a3 a4 + 4√b1 b2 b3 b4
4
4
4
√(a1 + b1 )(a2 + b2 )(a3 + b3 )( 3√a1 a2 a3 + 3√b1 b2 b3 ) ≥ √a1 a2 a3 3√a1 a2 a3 + √b1 b2 b3 3√b1 b2 b3
4
4
√(a1 + b1 )(a2 + b2 )(a3 + b3 ) √( 3√a1 a2 a3 + 3√b1 b2 b3 ) ≥ ( 3√a1 a2 a3 + 3√b1 b2 b3 )
4
4
3
√(a1 + b1 )(a2 + b2 )(a3 + b3 ) ≥ √( 3√a1 a2 a3 + √b1 b2 b3 )
3
3
√(a1 + b1 )(a2 + b2 )(a3 + b3 ) ≥ 3√a1 a2 a3 + 3√b1 b2 b3
Using CBS inequality to prove (1) is obviously tedious, but it can give a nice extension of this
inequality as we shall see.
My third attempt to deal with (1) is given below. I use “think backwards, write forwards” and the
reader may trace my way of thinking by beginning at the last line and read backwards.
By A.M.  G.M.,
a1
a2
a3
b1
b2
b3
+
+
+
+
a1 + b1 a 2 + b2 a 3 + b3 a1 + b1 a 2 + b2 a 3 + b3
+
3
3
3
≥√
3
⇔1≥
3
a1
a2
a3
b1
b2
b3
+√
a1 + b1 a 2 + b2 a 3 + b3
a1 + b1 a 2 + b2 a 3 + b3
3
√a1 a2 a3 + √b1 b2 b3
3
√(a1 +b1 )(a2 +b2 )(a3 +b3 )
3
⇔ √(a1 + b1 )(a2 + b2 )(a3 + b3 ) ≥ 3√a1 a2 a3 + 3√b1 b2 b3
Nice, short, isn’t it?
P.2
Last, but not the least, I try to give (1) a possible extension.
By studying
(1), (2) and (3), we can get:
n
n
n
n
n
∏(ai + bi ) ≥ √∏ ai + √∏ bi
i=1
Two possible proofs of
i=1
… . (𝟒)
i=1
(4) are :
(a) Use “Backward mathematical induction”. (a very long proof)
(b) Use my “third attempt”. (a very short proof)
I leave the proofs for the reader as an exercise. For those who don’t know much on backward
induction may find my example useful:
http://www.qc.edu.hk/math/Resource/AL/AGM_1.pdf
P.3