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# Inequality 3(George Basdekis)

Problem:

Let $\displaystyle a,b,c$ be positive real numbers. Prove that

$\displaystyle \frac{a}{bc}+\frac{1}{a}+\frac{b}{ca}+\frac{1}{b}+\frac{c}{ab}+\frac{1}{c}\geq \frac{1}{2}\left(\frac{a+b}{b^2+c^2}+\frac{b+c}{c^2+a^2}+\frac{c+a}{a^2+b^2}\right)$.

1st solution:

The left hand side can be rewritten as

$\displaystyle \frac{a^2+b^2+c^2+ab+bc+ca}{2abc}\geq \sum_{cyc}\frac{a+b}{b^2+c^2}$.

We only need to prove that

$\displaystyle \sum_{cyc}\frac{a^2+ab}{2abc}\geq \sum_{cyc}\frac{a+b}{b^2+c^2}$,

which is true according to the AM-GM inequality, that is

$\displaystyle b^2+c^2\geq 2bc\wedge c^2+a^2\geq 2ca \wedge a^2+b^2\geq 2ab$,

which holds for all non-negative numbers.

2nd solution:

Bringing everything in the left hand side we get that

$\displaystyle \frac{a^2+b^2+c^2+ab+bc+ca}{2abc}-\sum_{cyc}\frac{a+b}{b^2+c^2}\geq 0$.

But this one holds because it is of the form

$\displaystyle \sum_{cyc}\frac{a(a+b)(b-c)^{2}}{b^2+c^2}\geq 0$.

Equality occurs if and only if $\displaystyle a=b=c$, Q.E.D.