Matematické Fórum

If we divide the whole inequality by [mathjax]abc>0[/mathjax], we obtain

[mathjax] {\displaystyle \frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b} \geq \sqrt{\left(\frac{a}{c}+\frac{a}{b}\right)\left(1+\frac{a}{c}\right)} + \sqrt{\left(\frac{b}{a}+\frac{b}{c}\right)\left(1+\frac{b}{a}\right)} + \sqrt{\left(\frac{c}{b}+\frac{c}{a}\right)\left(1+\frac{c}{b}\right)}}.[/mathjax]

The left-hand side can be rearranged in the form

[mathjax] {\displaystyle \frac{1}{2}\left[\left(\frac{a}{c}+\frac{a}{b}\right) + \left(1+\frac{a}{c}\right)\right]+
   \frac{1}{2}\left[\left(\frac{b}{a}+\frac{b}{c}\right) + \left(1+\frac{b}{a}\right)\right] +
   \frac{1}{2}\left[\left(\frac{c}{b}+\frac{c}{a}\right) + \left(1+\frac{c}{b}\right)\right] + \frac{1}{2}\left(\frac{c}{a}+\frac{a}{b} + \frac{b}{c} -3 \right).}[/mathjax]

For each term in the square brackets, we use the AG-inequality [mathjax] x+y\geq 2\sqrt{xy} [/mathjax], whereas for the last term there holds

[mathjax] {\displaystyle \frac{c}{a}+\frac{a}{b} + \frac{b}{c} -3 \geq 3\sqrt[3]{\frac{c}{a}\cdot\frac{a}{b} \cdot \frac{b}{c}} -3 = 0,} [/mathjax]

where we have applied another AG-inequality [mathjax] x+y+z \geq 3\sqrt[3]{xyz}.[/mathjax]