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Algebra Difficulty 5.7 AIME, harder Prove it

5. Given x,y,z∈R+,x2+y2+z2≤3xyzx, y, z \in R^{+}, x^{2}+y^{2}+z^{2} \leq \sqrt{3} x y z, prove: x+y+z≤xyzx+y+z \leq x y z.

Solution

We can easily prove that ab+bc+ca≤13(a+b+c)2,ab+bc+ca≤a2+b2+c2a b+b c+c a \leq \frac{1}{3}(a+b+c)^{2}, a b+b c+c a \leq a^{2}+b^{2}+c^{2}, thus we have
x+y+zxyz=1xy+1yz+1zx≤13(1x+1y+1z)2=13(xy+yz+zxxyz)2≤13(x2+y2+z2xyz)2≤13(3xyzxyz)2=1\begin{array}{l} \frac{x+y+z}{x y z}=\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \leq \frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2} \\ =\frac{1}{3}\left(\frac{x y+y z+z x}{x y z}\right)^{2} \\ \leq \frac{1}{3}\left(\frac{x^{2}+y^{2}+z^{2}}{x y z}\right)^{2} \\ \leq \frac{1}{3}\left(\frac{\sqrt{3} x y z}{x y z}\right)^{2}=1 \end{array}

Therefore, x+y+z≤xyzx+y+z \leq x y z.

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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.