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[huyne123]

$x(x+1)+y(y+1)+z(z+1) \leq 18$ CM

$\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}$

[onpiece123]

Ta có 18$\geq x^{2}+y^{2}+z^{2}+x+y+z$ <=> 18$\geq \frac{(x+y+z)^{2}}{3}+(x+y+z)$ <=> x+y+z$\leq 6$

 Áp dụng bđt cauchy -schwarz ta có :

    $\frac{1}{x+y+1}+\frac{1}{x+z+1}+\frac{1}{y+z+1}\geq \frac{9}{2(x+y+z)+3}$$\geq \frac{3}{5}$ ( vì x+y+z$\leq 6$)

    Suy ra (đpcm)

  Dấu "=" xảy ra khi và chỉ khi x=y=z=2