設(shè)

一或三象限切點(diǎn)與原點(diǎn)連線斜率k1

二或四象限切點(diǎn)與原點(diǎn)連線斜率k2

有

k1＞0

k2=-b^4/(a^4k1)

即

S△AOB

=

b^4+a^4k12

/

2√((b^4+a^4k12)2/a2b2+(b2-a2）2k12)

=

1

/

2√(1/a2b2+(b2-a2）2k12/(b^4+a^4k12)2)

=

1

/

2√(1/a2b2+(b2-a2）2/(b^4/k1+a^4k1)2)

即

S△AOB

≥

1

/

2√(1/a2b2+(b2-a2）2/(4a^4b^4))

=

1

/

2√((b2+a2）2/(4a^4b^4))

=

1

/

2(b2+a2）/(2a2b2)

=

a2b2/(a2+b2)

且

S△AOB

≤

1

/

2√(1/a2b2)

=

ab/2

即

a2b2/(a2+b2)≤S△AOB≤ab/2

得證

設(shè)

x2/((a2+b2)/a2)+y2/((a2+b2)/b2)=1

上點(diǎn)與原點(diǎn)連線長(zhǎng)度d

有

S△APB

=

1/2·(d2-1)·2√(d2-1)/d2·ab

=

(d2-1)^(3/2)/d2·ab

設(shè)

t=d2-1

有

S△APB=ab·t^(3/2)/(t+1)，b2/a2≤t≤a2/b2

即

S△APB

≥

ab

b3/a3

/

(a2+b2)/a2

=

b^4/(a2+b2)

且

S△APB

≤

ab

a3/b3

/

(a2+b2)/b2

=

a^4/(a2+b2)

即

b^4/(a2+b2)≤S△APB≤a^4/(a2+b2)

得證