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Algebra Difficulty 7.2 National olympiad, round 2 Prove it

42. Given real numbers x,y,zx, y, z satisfying xyz=−1x y z = -1, prove that: x4+y4+z4+3(x+y+z)⩾x2y+x2z+y2x+y2z+z2x+z2yx^{4} + y^{4} + z^{4} + 3(x + y + z) \geqslant \frac{x^{2}}{y} + \frac{x^{2}}{z} + \frac{y^{2}}{x} + \frac{y^{2}}{z} + \frac{z^{2}}{x} + \frac{z^{2}}{y}. (2004 Iran Mathematical Olympiad Problem)

Solution

42. Since xyz=−1x y z=-1, then
x4+y4+z4+3(x+y+z)−(x2y+x2z+y2x+y2z+z2x+z2y)=x4+y4+z4−3(x+y+z)xyz−x2(y+z)yz−y2(z+x)zx−z2(x+y)xy=x4+y4+z4−3(x+y+z)xyz−x3(y+z)xyz−y3(z+x)zxy−z3(x+y)xyz=x4+y4+z4−3(x+y+z)xyz+x3(y+z)+y3(z+x)+z3(x+y)=x4+x3(y+z)+y4+y3(z+x)+z4+z3(x+y)−3(x+y+z)xyz=(x+y+z)(x3+y3+z3−3xyz)=(x+y+z)2(x2+y2+z2−xy−yz−zx)=12(x+y+z)2[(x−y)2+(y−z)2+(z−x)2]⩾0\begin{array}{l} x^{4}+y^{4}+z^{4}+3(x+y+z)-\left(\frac{x^{2}}{y}+\frac{x^{2}}{z}+\frac{y^{2}}{x}+\frac{y^{2}}{z}+\frac{z^{2}}{x}+\frac{z^{2}}{y}\right)= \\ x^{4}+y^{4}+z^{4}-3(x+y+z) x y z-\frac{x^{2}(y+z)}{y z}-\frac{y^{2}(z+x)}{z x}-\frac{z^{2}(x+y)}{x y}= \\ x^{4}+y^{4}+z^{4}-3(x+y+z) x y z-\frac{x^{3}(y+z)}{x y z}-\frac{y^{3}(z+x)}{z x y}-\frac{z^{3}(x+y)}{x y z}= \\ x^{4}+y^{4}+z^{4}-3(x+y+z) x y z+x^{3}(y+z)+y^{3}(z+x)+z^{3}(x+y)= \\ x^{4}+x^{3}(y+z)+y^{4}+y^{3}(z+x)+z^{4}+z^{3}(x+y)-3(x+y+z) x y z= \\ (x+y+z)\left(x^{3}+y^{3}+z^{3}-3 x y z\right)= \\ (x+y+z)^{2}\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)= \\ \frac{1}{2}(x+y+z)^{2}\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \geqslant 0 \end{array}

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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.