Maths Olympiad Prep

Library / /168 of 520

Algebra Difficulty 6.3 National olympiad Prove it

Example 3 (Self-created problem, 2000.07.26) Let x,y,zR+x, y, z \in \mathbf{R}^{+}, then
(xy+z)x2(x+y+z)92xyz\left(\sum \frac{x}{y+z}\right) \cdot \sum x^{2}(-x+y+z) \leqslant \frac{9}{2} x y z

Equality in (3) holds if and only if x=y=zx=y=z.

Solution

9xyz(y+z)2[x2(x+y+z)][x(x+y)(x+z)]=9xyzyz(y+z)+18x2y2z22{[x2(y+z)]2(x3)2}6xyzx2(x+y+z)=2[x6+3x2y2z2y2z2(y2+z2)]+xyz[2x3yz(y+z)]0\begin{array}{l}9 x y z \prod(y+z)-2\left[\sum x^{2}(-x+y+z)\right]\left[\sum x(x+y)(x+z)\right]= \\ 9 x y z \sum y z(y+z)+18 x^{2} y^{2} z^{2}-2\left\{\left[\sum x^{2}(y+z)\right]^{2}-\sum\left(x^{3}\right)^{2}\right\}- \\ 6 x y z \sum x^{2}(-x+y+z)= \\ 2\left[\sum x^{6}+3 x^{2} y^{2} z^{2}-\sum y^{2} z^{2}\left(y^{2}+z^{2}\right)\right]+ \\ x y z\left[2 \sum x^{3}-\sum y z(y+z)\right] \geqslant 0\end{array}

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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