562 Real Analysis
u=
8 v
v^2 + 3,
v=8 u
u^2 + 3.
This we transform into
uv^2 + 3 u= 8 v,
u^2 v+ 3 v= 8 u,then subtract the second equation from the first, to obtainuv(v−u)= 11 (v−u). Either
u=voruv=11. The first possibility impliesu=v =0oru=v =±
√
- The
second impliesuv^2 + 3 u= 11 v+ 3 u= 8 vsou=−v, which gives rise to the equality
u=−u (^28) +u 3. This can hold only whenu=0. The critical points off (x, y)are therefore
(
3
2
,
3
2
)
,
(
3 +
√
5
2
,
3 +
√
5
2
)
,
(
3 −
√
5
2
,
3 −
√
5
2
)
.
We computef(^32 ,^32 )= 165 andf(^3 ±
√ 5
2 ,3 ±√ 5
2 )=0.
What about the behavior offwhen|x|+|y|→∞? We compute thatf (x, y)=x^2 y^2 − 3 xy^2 − 3 x^2 y+ 5 xy+ 3 x^2 + 3 y^2 − 3 x− 3 y+ 1.Note that when|x|+|y|→∞,
1
2
x^2 +1
2
y^2 − 3 x− 3 y+ 1 →∞,
5
2x^2 +5
2
y^2 + 5 xy=5
2
(x+y)^2 ≥ 0 ,x^2 y^2 − 3 xy^2 − 3 x^2 y=x^2(
y−3
2
) 2
+y^2(
x−3
2
) 2
−
9
2
≥−
9
2
.
By adding these we deduce that when|x|+|y|→∞,f (x, y)→∞.
We conclude that 0 is the absolute minimum forf. This proves the inequality. And
as we just saw, equality is achieved when
a
b=
c
d=
3 +
√
5
2
or
a
b=
c
d=
3 −
√
5
2
.
(Mathematical Reflections, proposed by T. Andreescu)507.Consider a coordinate system in the plane and let thenpoints beP 1 (x 1 ,y 1 ),
P 2 (x 2 ,y 2 ),...,Pn(xn,yn). For an oriented linel, we will denote byl⊥the oriented