Maths Olympiad Prep

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Algebra Difficulty 7.1 National olympiad, round 2 Prove it

Question 4-2 Let x,y,zx, y, z be real numbers, and satisfy x2+y2+z2=r2x^{2}+y^{2}+z^{2}=r^{2} (r>0)(r>0), prove that: x+y+z−2r2xyz⩽2rx+y+z-\frac{2}{r^{2}} x y z \leqslant \sqrt{2} r.

For the letter rr in question 4-2, taking special values, we get:
1. (1991 Polish Mathematical Competition) Let x,y,zx, y, z be real numbers, and satisfy x2+y2+z2=1x^{2}+y^{2}+z^{2}=1, prove that: x+y+z−2xyz⩽2x+y+z-2 x y z \leqslant \sqrt{2}.
2. (2008 High School Mathematics League Sprint Question) Let x,y,zx, y, z be real numbers, and satisfy x2+y2+z2=2x^{2}+y^{2}+z^{2}=2, prove that: x+y+z−xyz⩽2x+y+z-x y z \leqslant 2.

Question 5-1 (2009 Northern Mathematical Competition) If x,y,z>0x, y, z>0, and x2+y2+z2=3x^{2}+y^{2}+z^{2}=3, prove that: x2009−2008(x−1)y+z+y2009−2008(y−1)z+x+z2009−2008(z−1)x+y⩾12(x+y+z)\frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y} \geqslant \frac{1}{2}(x+y+z).

Solution

Prove using the 2009-term AM-GM inequality, we get
x2009+2008=x2009+12009+12009+⋯+12009⩾2009x, i.e., x2009−2008(x−1)⩾x.\begin{array}{l} x^{2009}+2008=x^{2009}+1^{2009}+1^{2009}+\cdots+1^{2009} \geqslant 2009 x, \\ \text { i.e., } x^{2009}-2008(x-1) \geqslant x . \end{array}

Similarly, y2009−2008(y−1)⩾yy^{2009}-2008(y-1) \geqslant y,
z2009−2008(z−1)⩾zz^{2009}-2008(z-1) \geqslant z

Noting the common inequality −xy+z+yz+x+zx+y⩾32-\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac{3}{2}, we get
x2009−2008(x−1)y+z+y2099−2008(y−1)z+x+z2009−2008(z−1)x+y⩾xy+z+yz+x+zx+y⩾32=14(x2+y2+z2+3)=14[(x2+1)+(y2+1)+(z2+1)]⩾14[2x+2y+2z]=x+y+z2\begin{array}{l} \frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2099}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y} \\ \geqslant \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac{3}{2}=\frac{1}{4}\left(x^{2}+y^{2}+z^{2}+3\right) \\ =\frac{1}{4}\left[\left(x^{2}+1\right)+\left(y^{2}+1\right)+\left(z^{2}+1\right)\right] \\ \geqslant \frac{1}{4}[2 x+2 y+2 z]=\frac{x+y+z}{2} \end{array}

Thus, the original inequality is proved.
By changing the year in the problem to a general letter, we get

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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.