You can rewrite the equation
a.B + b.C + c.A as
assuming that the positive numbers are integers and only hold one place value. A rational fraction or an irrational number wouldn't make sense if you mean to use ' . ' as a decimal point
Follows that this is equivalent to
Try solving from here.
I'm not sure what you mean by "only with using properties of linear function". But here is a proof that uses only elementary algebra.
I'll start with some apparently different inequalities.
Step 1. If x ≤ 1 and y ≤ 1 then x + y ≤ 1 + xy.
Proof. Follows from the fact that (1 – x)(1 – y) ≥ 0.
Step 2. If –1 ≤ x ≤ 1, –1 ≤ y ≤ 1 and –1 ≤ z ≤ 1, then yz + zx + xy + 1 ≥0.
Proof. The result clearly holds if x, y and z are all positive, or all negative. (It also holds if any of them are zero.) So suppose that x and y have one sign, and z has the opposite sign. In fact, changing the sign of all three variables (which does not change the value of yz + zx + xy + 1), we may assume that x and y are positive and that z is negative, say z = –w, with 0 ≤ w ≤ 1. Then yz + zx + xy + 1 = 1 + xy – w(x+y) ≥ 1 + xy – (x+y) ≥ 0, by step 1.
Step 3. Now suppose that a, A, b, B, c, C are as in the statement of the problem. Since A = s–a is nonnegative, as is a, it follows that
0 ≤ a ≤ s, and similarly for b and c. Therefore lies between –1 and 1, with similar results for b and c.
Step 4. Now apply step 2, with . It follows that . Multiply out the brackets and simplify this, to get .
Step 5. The original problem was to find an upper bound for aB + bC + cA. But this is equal to a(s–b) + b(s–c) + c(s–a) = s(a+b+c) – (bc+ca+ab). So the result follows from step 4.
[The inequality need not be strict. It becomes an equality, for example, if a=b=0 and c=s.]