2.1 Identities and Inequalities 41Then we apply the AM–GM inequality to the last term in each denominator to obtain the
stronger inequality
4 a
a+ 5 (b+c+d)+
4 b
b+ 5 (c+d+a)+
4 c
c+ 5 (d+a+b)+
4 d
d+ 5 (a+b+c)≥ 1 ,
which we proceed to prove.
In order to simplify computations, it is better to denote the four denominators by
16 x, 16 y, 16 z, 16 w, respectively. Thena+b+c+d = x+y+z+w, and so
4 a+ 16 x= 4 b+ 16 y= 4 c+ 16 z= 4 d+ 16 w= 5 (x+y+z+ 2 ). The inequality
becomes
− 11 x+ 5 (y+z+w)
16 x+
− 11 y+ 5 (z+w+x)
16 y+
− 11 z+ 5 (w+x+y)
16 z
+
− 11 w+ 5 (x+y+z)
16 w≥ 1 ,
or− 4 · 11 + 5(
y
x+
z
x+
w
x+
z
y+
w
y+
x
y+
w
z+
x
z+
y
z+
x
w+
y
w+
z
w)
≥ 16.
And this follows by applying the AM–GM inequality to the twelve summands in the
parentheses.
Try your hand at the following problems.121.Show that all real roots of the polynomialP(x)=x^5 − 10 x+35 are negative.
122.Prove that for any positive integern,nn− 1 ≥nn+ 21
(n− 1 ).123.Leta 1 ,a 2 ,...,anandb 1 ,b 2 ,...,bnbe nonnegative numbers. Show that(a 1 a 2 ···an)^1 /n+(b 1 b 2 ···bn)^1 /n≤((a 1 +b 1 )(a 2 +b 2 )···(an+bn))^1 /n.124.Leta, b, cbe the side lengths of a triangle with semiperimeter 1. Prove that1 <ab+bc+ca−abc≤28
27
.
125.Which number is larger,∏^25n= 1(
1 −
n
365)
or