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a,b,c 為三角行三邊長, T 為三角行面積
prove that: aa+bb+cc >= 4*sqrt(3)*T

One approach is a routine slog from Heron's formula. The inequality is quickly shown to be equivalent to a2b2 + b2c2 + c2a2 <= a4 + b4 + c4, which is true since a2b2 <= (a4 + b4)/2. We get equality iff the triangle is equilateral.

Another approach is to take an altitude lying inside the triangle. If it has length h and divides the base into lengths r and s, then we quickly find that the inequality is equivalent to (h - (r + s)Ö3/2)2 + (r - s)2 >= 0, which is true. We have equality iff r = s and h = (r + s)Ö3/2, which means the triangle is equilateral.

A third solution is due to Jonathan Mizrahi (somewhat adapted):

We have b2 + c2 >= 2bc with equality iff b = c. Also for any angle x in the range 0o to 180o we have 2bc >= 2bc sin(X + 30o) with equality iff X = 60o. So taking X to be the angle between the sides b and c (we cannot call it A because A is already used to mean the area in this question!) we have that b2 + c2 >= bc sin(X + 30o) with equality iff the triangle is equilateral. Now 2 sin(X + 30o) = Ö3 sin X + cos X, so using the cosine rule a2 = b2 + c2 - 2bc cos X, we get the required inequality.

IMO1961 Q2
http://www.kalva.demon.co.uk/imo/isoln/isoln612.html

prove that: aa+bb+cc >= 4*sqrt(3)*T

s=(a+b+c)/2

(s-a)(s-b)(s-c)<=[(s-a+s-b+s-c)/3]^3
=(s/3)^3

4sqrt3 T
<=4sqrt3 sqrt[s^4/3^3]
=4sqrt3 s^2/3sqrt3
=(a+b+c)^2/3
<=a^2+b^2+c^2 (by cauchyscwharz inequality)

am i right?