Olympiad Maths Prep

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Problem 1044

National olympiad, first round
Algebra Difficulty 6.0 Prove it

17. Let x,y,z∈R+,x+y+z=1,nx, y, z \in \mathbf{R}_{+}, x+y+z=1, n be a positive integer, then
x4y(1−yn)+y1z(1−zn)+z4x(1−xn)⩾3n3n+2−9. \frac{x^{4}}{y\left(1-y^{n}\right)}+\frac{y^{1}}{z\left(1-z^{n}\right)}+\frac{z^{4}}{x\left(1-x^{n}\right)} \geqslant \frac{3^{n}}{3^{n+2}-9} .

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

17. Apply the inequality
xn+1+yn+1+zn+1⩾(x+y+z)n+13x^{n+1}+y^{n+1}+z^{n+1} \geqslant \frac{(x+y+z)^{n+1}}{3} and the given condition x+y+z=1x+y+z=1.
 The left side of the inequality =(x2)2y−yn+1+(y2)2z−zn+1+(z2)2x−xn+1⩾(x2+y2+z2)2(x+y+z)−(xn+1+yn+1+zn+1)⩾(13)21−13n=3n3n+2−9. \begin{aligned} \text { The left side of the inequality } & =\frac{\left(x^{2}\right)^{2}}{y-y^{n+1}}+\frac{\left(y^{2}\right)^{2}}{z-z^{n+1}}+\frac{\left(z^{2}\right)^{2}}{x-x^{n+1}} \geqslant \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{(x+y+z)-\left(x^{n+1}+y^{n+1}+z^{n+1}\right)} \\ & \geqslant \frac{\left(\frac{1}{3}\right)^{2}}{1-\frac{1}{3^{n}}}=\frac{3^{n}}{3^{n+2}-9} . \end{aligned}

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