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a^2 + b^2 + c^2 <= (k^2 + 1)(ab + bc + ca) if a/b+b/c+ c/a= (k + 1)^2 + 2/(k+1)

Source: 2015 VMEO IV Seniors 11.1 Vietnamese Mathematics e - Olympiad https://artofproblemsolving.com/community/c2463156_2015_vmeo_iv

September 19, 2021
algebrainequalities

Problem Statement

Let k≥0k \ge 0 and a,b,ca, b, c be three positive real numbers such that ab+bc+ca=(k+1)2+2k+1.\frac{a}{b}+\frac{b}{c}+ \frac{c}{a}= (k + 1)^2 + \frac{2}{k+ 1}. Prove that a2+b2+c2≤(k2+1)(ab+bc+ca).a^2 + b^2 + c^2 \le (k^2 + 1)(ab + bc + ca).
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