34 2 Algebra
Example.Find the maximum of the functionf (x, y, z)= 5 x− 6 y+ 7 zon the ellipsoid
2 x^2 + 3 y^2 + 4 z^2 =1.
Solution.For a point(x,y,z)on the ellipsoid,
(f (x, y, z))^2 =( 5 x− 6 y+ 7 z)^2 =(
5
√
2
·
√
2 x−6
√
3
·
√
3 y+7
2
· 2 z) 2
≤
((
5
√
2
) 2
+
(
−
6
√
3
) 2
+
(
7
2
) 2 )(
(
√
2 x)^2 +(√
3 y)^2 +( 2 z)^2)
=
147
4
( 2 x^2 + 3 y^2 + 4 z^2 )=147
4
.
Hence the maximum offis
√
147 /2, reached at the point(x,y,z)on the ellipsoid for
whichx, z >0,y<0, andx:y:z=√^52 :−√^63 :^72.
The next problem was on the short list of the 1993 International Mathematical
Olympiad, being proposed by the second author of the book.
Example.Prove that
a
b+ 2 c+ 3 d+
b
c+ 2 d+ 3 a+
c
b+ 2 a+ 3 b+
d
a+ 2 b+ 3 c≥
2
3
,
for alla, b, c, d >0.
Solution.Denote byEthe expression on the left. Then
4 (ab+ac+ad+bc+bd+cd)E
=(a(b+ 2 c+ 3 d)+b(c+ 2 d+ 3 a)+c(d+ 2 a+ 3 b)+d(a+ 2 b+ 3 c))×(
a
b+ 2 c+ 3 d+
b
c+ 2 d+ 3 a+
c
b+ 2 a+ 3 b+
d
a+ 2 b+ 3 c)
≥(a+b+c+d)^2 ,where the last inequality is a well-disguised Cauchy–Schwarz. Finally,
3 (a+b+c+d)^2 ≥ 8 (ab+ac+ad+bc+bd+cd),because it reduces to
(a−b)^2 +(a−c)^2 +(a−d)^2 +(b−c)^2 +(b−d)^2 +(c−d)^2 ≥ 0.Combining these two and cancelling the factorab+ac+ad+bc+bd+cd, we obtain
the inequality from the statement.