Silk Road Mathematics Competition

Let $x$, $y$, $z$ be positive real numbers such that $a=x/y$, $b=y/z$, $c=z/x$. After substitution and simplifying the given inequality is transformed to the following one: $$4(x+y+z{)}^{\sqrt{3}xyz}\le 3(x+y{)}^{2/3}(y+z{)}^{2/3}(z+x{)}^{2/3}$$ For triangle with the sides lengths $u=x+y$, $v=y+z$, $w=z+x$ the last inequality gives $$4p\sqrt[3]{(p-u)(p-v)(p-w)}\le 3(uvw{)}^{2/3}$$ where $p$ is a semiperimeter of the triangle. By Heron's formula we obtain that the inequality is equivalent to: $$(2p{)}^{2/3}\le (3\sqrt{3}\frac{uvw}{4s}{)}^{2/3}=(3\sqrt{3}R{)}^{2/3}$$ where $S$ is area and $R$ is a radius of the circumcircle of the triangle. But the inequality $$u+v+w=2p\le 3\sqrt{3}R$$ is well-known; there are several ways to prove that.

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Alternative solution.

By raising to the third power the given inequality and taking into account the condition $abc=1$ we eliminate radicals. After simplifying we get the following inequality: $$114+30\sum _{cyc}\left(a+\frac{1}{a}\right)+10\sum _{cyc}\frac{a}{b}\le 54\sum _{cyc}\frac{b}{a}+27\sum _{cyc}\left(a^2+\frac{1}{a^2}\right)$$ The last inequality can be obtained from the following elementary inequalities: $$2\sum _{cyc}\frac{a}{b}\le \sum _{cyc}\left(a^2+\frac{1}{a^2}\right),3\le \sum _{cyc}\frac{b}{a},$$ $$a+\frac{1}{a}\le \frac{1}{4}\left(a^2+\frac{1}{a^2}+6\right),2\le a^2+\frac{1}{a^2}.$$