If one of a = one of B, a = one of C, B = one of 3, then a, B and C are ()

If one of a = one of B, a = one of C, B = one of 3, then a, B and C are ()

Because 1 / a = 1 / 3, so a = 3
1 / a = A / B, substituting a = 3 into b = 9
A / b = B / C, substitute a = 3, B = 9 into C = 27

Given A-B = B-C = 3 / 5 and a + B + C = 1, we can find the value of Ba + Ca + BC

So (a-b) ^ 2 + (B-C) ^ 2 + (C-A) ^ 2 = (3 / 5) ^ 2 + (3 / 5) ^ 2 + (6 / 5) ^ 2 = 54 / 25, so a ^ 2-2ab + B ^ 2 + B ^ 2-2bc + C ^ 2 + a ^ 2-2ac + C ^ 2 = 54 / 252 (a ^ 2 + B ^ 2 + C ^ 2) - 2 (Ba + Ca + BC) = 54 / 25ba + Ca + BC = (2-54 / 25) / 2

Given 2 / a = 3 / b = 4 / C, find AB + BC + AC of a, B and C

Let a / 2 = B / 3 = C / 4 = R, then a = 2R, B = 3R and C = 4R
A²/(AB+BC+AC)=4R²/（6R²+12R²+8R²）=2/13
B²/(AB+BC+AC)=9/26
C²/(AB+BC+AC)=8/13

It is known that a, B, C ∈ R, a ^ 2, B ^ 2, C ^ 2 = 1
It is known that a, B, C ∈ R, a ^ 2 + B ^ 2 + C ^ 2 = 1

Because a ^ 2 + B ^ 2 > = 2Ab note: from (a-b) ^ 2 > = 0
Similarly, B ^ 2 + C ^ 2 > = 2BC
a^2+c^2>=2ac
To prove | a + B + C ≤ √ 3 is to prove (a + B + C) ^ 2 ≤ 3
(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc = 1+2ab+2ac+2bc ≤ 1+(a^2+b^2)+(b^2+c^2)+(a^2+c^2)=
1+2(a^2+b^2+c^2)=3
So (a + B + C) ^ 2 ≤ 3, so we prove that if and only if a = b = C, we take the equal sign