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Algebra Difficulty 7.6 National olympiad, round 2 Prove it Turkey

Show that for all positive real numbers xx, yy and zz
x2+1(x+y)2+4(z+1)+y2+1(y+z)2+4(x+1)+z2+1(z+x)2+4(y+1)≥12 \frac{x^2+1}{(x+y)^2+4(z+1)} + \frac{y^2+1}{(y+z)^2+4(x+1)} + \frac{z^2+1}{(z+x)^2+4(y+1)} \ge \frac{1}{2}

Solution

Note that (x+y)2≤2(x2+y2)(x+y)^2 \le 2(x^2+y^2) and 4z+4≤2(z2+3)4z+4 \le 2(z^2+3). Therefore,
x2+1(x+y)2+4(z+1)≥x2+12(x2+y2+z2+3) \frac{x^2+1}{(x+y)^2+4(z+1)} \ge \frac{x^2+1}{2(x^2+y^2+z^2+3)}
Similarly we can write analogous inequalities for pairs (y,z)(y, z) and (z,x)(z, x). The sum of these three inequalities yields
x2+1(x+y)2+4(z+1)+y2+1(y+z)2+4(x+1)+z2+1(z+x)2+4(y+1)≥x2+y2+z2+32(x2+y2+z2+3)=12. \frac{x^2+1}{(x+y)^2+4(z+1)} + \frac{y^2+1}{(y+z)^2+4(x+1)} + \frac{z^2+1}{(z+x)^2+4(y+1)} \\ \ge \frac{x^2+y^2+z^2+3}{2(x^2+y^2+z^2+3)} = \frac{1}{2}.

The equality holds at x=y=z=1x = y = z = 1.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.