Let be real numbers such that Prove that Where

Here's the proof of the first problem:

By the triangle inequality, we have

|a-kb|+|b-kc|+|c-ka| ≤ |a|+|b|+|c| +k(|a|+|b|+|c|)=(1+k)(|a|+|b|+|c|).

Moreover, if a^2 < b^2, then a<b, given that both a and b are positive. Hence, all we need to show is that the inequality holds for the squares of the given expressions. Thus, we have

(1+k)^2(|a|+|b|+|c|)^2=(1+k)^2(a^2+b^2+c^2+|ab|+|bc|+|ca|)

We are given that a^2+b^2+c^2=1. Hence,

ab+bc+ca ≤ (a^2+b^2)/2 + (b^2+c^2)/2 + (c^2+a^2)/2 = a^2+b^2+c^2=1,

And since this holds for all a, b, c, it surely holds if we substitute a by |a|, b by |b|, and c by |c|.

Thus,

(1+k)^2(a^2+b^2+c^2+|ab|+|bc|+|ca|) ≤ 2(1+k)^2=2k^2+4k+2.

Furthermore, we have

(sqrt(3k^2 + 2k + 3))^2=3k^2 + 2k + 3.

Lastly, 2k^2+4k+2 ≤ 3k^2+2k+3 for all k ≥ 0.

Therefore, the inequality

|a-kb|+|b-kc|+|c-ka| ≤ sqrt(3k^2+2k+3) holds for all k ≥ 0, as required.