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Algebra Difficulty 7.5 National olympiad, round 2 Prove it

14. Let x1,x2,x3,x4,x5,x6x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6} all be positive numbers, prove that:
x1x2+x3+x2x3+x4+x3x4+x5+x4x5+x6+x5x6+x1+x6x1+x2⩾3\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{4}}+\frac{x_{3}}{x_{4}+x_{5}}+\frac{x_{4}}{x_{5}+x_{6}}+\frac{x_{5}}{x_{6}+x_{1}}+\frac{x_{6}}{x_{1}+x_{2}} \geqslant 3

Solution

14.
(x1x2+x3+x2x3+x4+x3x4+x5+x4x5+x6+x5x6+x1+x6x1+x2)[x1(x2+x3)=+x2(x3+x4)+x3(x4+x5)+x4(x5+x6)+x5(x6+x1)+x6(x1+x2)=]⩾(x5+x2+x3+x4+x5+x6)2\begin{array}{l} \left(\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{4}}+\frac{x_{3}}{x_{4}+x_{5}}+\frac{x_{4}}{x_{5}+x_{6}}+\frac{x_{5}}{x_{6}+x_{1}}+\frac{x_{6}}{x_{1}+x_{2}}\right) \\ {\left[x_{1}\left(x_{2}+x_{3}\right)=+x_{2}\left(x_{3}+x_{4}\right)+x_{3}\left(x_{4}+x_{5}\right)+\right.} \\ \left.x_{4}\left(x_{5}+x_{6}\right)+x_{5}\left(x_{6}+x_{1}\right)+x_{6}\left(x_{1}+x_{2}\right)=\right] \geqslant \\ \left(x_{5}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)^{2} \end{array}

Additionally, since
(x1+x2+x3+x4+x5+x6)2−3[x1(x2+x3)+x2(x3+x4)+x3(x4+x5)+x4(x5+x6)+x5(x6+x1)+x6(x1+x2)]=(x1+x4)2+(x2+x5)2+(x3+x6)2(x1x2+x1x3+x2x3+x2x4+x3x4+x3x5+x4x5+x4x6+x5x6+x5x1+x6x1+x6x2)=12[(x1+x4−x2−x5)2+(x2+x5−x3−x6)2+(x3+x6−x1−x4)2]⩾0\begin{array}{l} \left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)^{2}-3\left[x_{1}\left(x_{2}+x_{3}\right)+x_{2}\left(x_{3}+x_{4}\right)+x_{3}\left(x_{4}+x_{5}\right)+\right. \\ \left.x_{4}\left(x_{5}+x_{6}\right)+x_{5}\left(x_{6}+x_{1}\right)+x_{6}\left(x_{1}+x_{2}\right)\right]= \\ \left(x_{1}+x_{4}\right)^{2}+\left(x_{2}+x_{5}\right)^{2}+\left(x_{3}+x_{6}\right)^{2} \\ \left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}+x_{3} x_{5}+\right. \\ \left.x_{4} x_{5}+x_{4} x_{6}+x_{5} x_{6}+x_{5} x_{1}+x_{6} x_{1}+x_{6} x_{2}\right)= \\ \frac{1}{2}\left[\left(x_{1}+x_{4}-x_{2}-x_{5}\right)^{2}+\right. \\ \left.\left(x_{2}+x_{5}-x_{3}-x_{6}\right)^{2}+\left(x_{3}+x_{6}-x_{1}-x_{4}\right)^{2}\right] \geqslant 0 \end{array}

That is,
3[x1(x2+x3)+x2(x3+x4)+x3(x4+x5)+x4(x5+x6)+x5(x6+x1)+x6(x1+x2)]⩽(x1+x2+x3+x4+x5+x6)2\begin{array}{l} 3\left[x_{1}\left(x_{2}+x_{3}\right)+x_{2}\left(x_{3}+x_{4}\right)+x_{3}\left(x_{4}+x_{5}\right)+x_{4}\left(x_{5}+x_{6}\right)+\right. \\ \left.x_{5}\left(x_{6}+x_{1}\right)+x_{6}\left(x_{1}+x_{2}\right)\right] \leqslant\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)^{2} \end{array}

Therefore,
x1x2+x3+x2x3+x4+x3x4+x5+x4x5+x6+x5x6+x1+x6x1+x2⩾3\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{4}}+\frac{x_{3}}{x_{4}+x_{5}}+\frac{x_{4}}{x_{5}+x_{6}}+\frac{x_{5}}{x_{6}+x_{1}}+\frac{x_{6}}{x_{1}+x_{2}} \geqslant 3

From the above proof, it is known that equality holds if and only if x1=x2=x3=x4=x5=x6x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=x_{6}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.