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An interesting inequality

\[\large \frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{1}{3}\geq\frac{8}{9}\bigg(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\bigg)\] Let \(a,b,c>0\), prove the above inequality.


This problem was taken from a discussion on VMF (Viet Maths Forum)

Note by Gurīdo Cuong
3 months, 2 weeks ago

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Here's the solution. First, without losing generality, we can assume that \(ab+bc+ac=3\), the problem'll become \[a^2+b^2+c^2+1\geq\frac{8}{3}\bigg(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg)\] Now we'll do some manipulation on the variables in \(LHS\) \[\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a(a+b)(a+c)+b(b+a)(b+c)+c(c+a)(c+b)}{(a+b)(b+c)(c+a)}\] \[=\frac{a^3+b^3+c^3+(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)-abc}=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca+ab+bc+ca)+3abc}{3(a+b+c)-abc}\] \[=\frac{(a+b+c)(a^2+b^2+c^2)+3abc}{3(a+b+c)-abc}\] So now the inequality is \[a^2+b^2+c^2+1\geq\frac{8[(a+b+c)(a^2+b^2+c^2)+3abc]}{9(a+b+c)-3abc}\] \[\Rightarrow [9(a+b+c)-3abc](a^2+b^2+c^2+1)\geq 8(a+b+c)(a^2+b^2+c^2)+24abc\] \[\Leftrightarrow 9(a+b+c)(a^2+b^2+c^2)+9(a+b+c)-3abc(a^2+b^2+c^2)-3abc\geq 8(a+b+c)(a^2+b^2+c^2)+24abc\] \[\Leftrightarrow (a+b+c)(a^2+b^2+c^2+9)\geq 3abc(a^2+b^2+c^2+9)\] \[\Rightarrow a+b+c\geq 3abc\] The last inequality can be proven easily by these following inequalities: \((a+b+c)^2\geq 3(ab+bc+ca)\) and \(abc\leq\sqrt{\big(\frac{ab+bc+ca}{3}\big)^3}\). The equality holds when \(a=b=c\) Gurīdo Cuong · 3 months, 2 weeks ago

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