European Mathematical Cup

Rewrite the left hand side of inequality in following way: $$\frac{a+b+c+3}{4}=\frac{a+b+c+3}{4\sqrt{abc}}=\frac{a+1}{4\sqrt{abc}}+\frac{b+1}{4\sqrt{abc}}+\frac{c+1}{4\sqrt{abc}}$$ Rewrite denominators: $$\frac{a+1}{4\sqrt{abc}}+\frac{b+1}{4\sqrt{abc}}+\frac{c+1}{4\sqrt{abc}}=\frac{a+1}{2\sqrt{ab\cdot c}+2\sqrt{ac\cdot b}}+\frac{b+1}{2\sqrt{ba\cdot c}+2\sqrt{bc\cdot a}}+\frac{c+1}{2\sqrt{ca\cdot b}+2\sqrt{cb\cdot a}}=$$ and then by arithmetic mean-geometric mean inequality, we have $$\begin{array}{rl}& =\frac{a+1}{ab+c+ac+b}+\frac{b+1}{bc+a+ba+c}+\frac{c+1}{ca+b+cb+a}=\frac{a+1}{(a+1)(b+c)}+\frac{b+1}{(b+1)(a+c)}+\frac{c+1}{(c+1)(b+a)}=\\ & =\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{b+a}=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}.\end{array}$$

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Alternative solution.

We introduce change of variables: $x=a^3,y=b^3,z=c^3$. We now have the condition $xyz=1$. We apply Schur inequality (with exponent $r=1$) to the numerator of the left hand side: $$x^3+y^3+z^3+3xyz\ge x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y$$ to obtain inequality $$\frac{x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y}{4}\ge \frac{1}{x^3+y^3}+\frac{1}{y^3+z^3}+\frac{1}{z^3+x^3}.$$ We apply arithmetic mean-geometric mean inequality for the denominators of the right hand side: $$x^3+y^3\ge 2x^{3/2}{y}^{3/2}\Rightarrow \frac{1}{x^3+y^3}\le \frac{1}{2x^{3/2}{y}^{3/2}}=\frac{1}{2}{z}^2\sqrt{xy}$$ and similarly to the other terms. We now have to prove $$\frac{x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y}{4}\ge \frac{1}{2}{x}^2\sqrt{yz}+\frac{1}{2}{y}^2\sqrt{xz}+\frac{1}{2}{z}^2\sqrt{xy}$$ $$\frac{x^{2}y+x^{2}z+y^{2}x+y^{2}z+z^{2}x+z^{2}y}{2}\ge x^2\sqrt{yz}+y^2\sqrt{xz}+z^2\sqrt{xy}.$$ We apply arithmetic mean-geometric mean inequality in pairs on the left hand side: $$\frac{x^{2}y+x^{2}z}{2}\ge x^2\sqrt{yz}$$ $$\frac{y^{2}z+y^{2}x}{2}\ge y^2\sqrt{xz}$$ $$\frac{z^{2}x+z^{2}y}{2}\ge z^2\sqrt{xy}.$$ Summing up inequalities from above finishes the proof.