triangle area inequality
來源: haha2000 於 06-12-04 13:05:58
This is an old problem I knew many years ago. I hope what is in my memory is still good :).
T1 is a triangle with side length, a1, b1, and c1.
T2 is a triangle with side length, a2, b2, and c2.
T3 is a triangle with side length, a1+a2, b1+b2, and c1+c2. (It is not hard to show that: a1+a2, b1+b2, and c1+c2 can form a triangle.)
There is an interesting inequity among areas of the three triangles.
Let
S1 denote the area of T1,
S2 denote the area of T2, and
S3 denote the area of T3,
prove the following:
sqrt(S3) >= sqrt(S1) + sqrt(S2).
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Deuss:
use p, q, r, s to name the ratios between two triangles of the factors inside sqrt. we need to prove
((1+p)(1+q)(1+r)(1+s))^{1/4} >=
1+(pqrs)^{1/4}
Let U=(pqrs)^{1/4}, the above is equivalent to
(1+p)(1+q)(1+r)(1+s)>=1+4U+6U^2+4U^3+U4
since
p+q+r+s>=4U
pq+pr+ps+qr+qs+rs>=6U^2
pqr+pqs+prs+qrs>=4U^3
the inequality follows. The equal sign holds when p=q=r=s, i.e. if the two tranigles are similar.
