Source: V International Festival of Young Mathematicians Sozopol 2014, Theme for 10-12 grade
Tags: algebra, Inequality, triangle inequality, inequalities
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Prove that, if $a,b,c$ are sides of a triangle, then we have the following inequality:
$3(a^3 b+b^3 c+c^3 a)+2(ab^3+bc^3+ca^3 )\geq 5(a^2 b^2+a^2 c^2+b^2 c^2 )$.
$ \sum_{cyc}{a^3b + b^3a} >= 2 \sum_{cyc}{a^2b^2} \ $ ( by AM-GM inequality )
$$ \implies 3 \left( \sum_{cyc}{{a^3}b}\right) + 2
\left ( \sum_{cyc}{{b^3}a}\right) $$$$\implies \frac{5}{2} \left ( \sum_{cyc}{ a^3b +b^3 a} \right) + \frac{1}{2} \left (\sum_{cyc}{a^3b - b^3a}\right) >= 5 \left (\sum_{cyc}{a^2b^2} \right) + \frac{1}{2} \left (\sum_{cyc}{a^3b - b^3a}\right) $$To show : $ \left( \sum_{cyc}{a^3b - b^3a}\right) >= 0 $
$$ \iff \sum_{cyc}{a^3(b-c)} >= 0 $$Now in any $\Delta ABC$ we have $ a > b - c $
$$\implies \sum_{cyc}{a^3(b-c)} > \sum_{cyc}{ a^2(b-c)^2 } >= 0 $$$$\therefore 3 \left( \sum_{cyc}{{a^3}b}\right) + 2 \left ( \sum_{cyc}{{b^3}a}\right) >= 5 \left (\sum_{cyc}{a^2b^2} \right) $$HENCE PROVED
Equality will happen iff $\triangle ABC $ is equilateral