Maths Olympiad Prep

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Problem 824

AMC 10/12, early questions
Algebra Difficulty 3.8 Prove it Junior Macedonian Mathematical Olympiad · North Macedonia

Let xx, yy, zz be positive real numbers. Prove that xyx2+y2+2z2+yzy2+z2+2x2+zxz2+x2+2y2≤32. \sqrt{\frac{xy}{x^2 + y^2 + 2z^2}} + \sqrt{\frac{yz}{y^2 + z^2 + 2x^2}} + \sqrt{\frac{zx}{z^2 + x^2 + 2y^2}} \le \frac{3}{2}.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

We have
xyx2+y2+2z2+yzy2+z2+2x2+zxz2+x2+2y2≤xyxy+yz+zx+z2+yzxy+yz+zx+x2+zxxy+yz+zx+y2=xy(z+x)(y+z)+yz(x+y)(z+x)+zx(y+z)(x+y)≤xz+x+yy+z2+yx+y+zz+x2+zy+z+xx+y2=x+yx+y+y+zy+z+z+xz+x2=32. \begin{aligned} & \sqrt{\frac{xy}{x^2 + y^2 + 2z^2}} + \sqrt{\frac{yz}{y^2 + z^2 + 2x^2}} + \sqrt{\frac{zx}{z^2 + x^2 + 2y^2}} \le \\ & \sqrt{\frac{xy}{xy + yz + zx + z^2}} + \sqrt{\frac{yz}{xy + yz + zx + x^2}} + \sqrt{\frac{zx}{xy + yz + zx + y^2}} = \\ & \sqrt{\frac{xy}{(z + x)(y + z)}} + \sqrt{\frac{yz}{(x + y)(z + x)}} + \sqrt{\frac{zx}{(y + z)(x + y)}} \le \\ & \frac{\frac{x}{z + x} + \frac{y}{y + z}}{2} + \frac{\frac{y}{x + y} + \frac{z}{z + x}}{2} + \frac{\frac{z}{y + z} + \frac{x}{x + y}}{2} = \\ & \frac{\frac{x + y}{x + y} + \frac{y + z}{y + z} + \frac{z + x}{z + x}}{2} = \frac{3}{2}. \end{aligned} Equality holds if and only if x=y=zx = y = z.

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