Đáp án:
Giải thích các bước giải:
$\dfrac{1}{x(x+1)}+\dfrac{x}{2}+\dfrac{x+1}{4} \geq 3\sqrt[3]{\dfrac{x(x+1)}{8x(x+1)}}=\dfrac{3}{2}$
Tương tự:
$\dfrac{1}{y(y+1)}+\dfrac{y}{2}+\dfrac{y+1}{4} \geq \dfrac{3}{2}$
$\dfrac{1}{z(z+1)}+\dfrac{z}{2}+\dfrac{z+1}{4} \geq \dfrac{3}{2}$
Cộng vế với vế:
$⇒\dfrac{1}{x^2+x}+\dfrac{1}{y^2+y}+\dfrac{1}{z^2+z}+\dfrac{x+y+z}{2}+\dfrac{x+y+z+3}{4} \geq \dfrac{9}{2}$
$⇒\dfrac{1}{x^2+x}+\dfrac{1}{y^2+y}+\dfrac{1}{z^2+z}+\dfrac{3}{2}+\dfrac{6}{4} \geq \dfrac{9}{2}$
$⇒\dfrac{1}{x^2+x}+\dfrac{1}{y^2+y}+\dfrac{1}{z^2+z} \geq \dfrac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $x=y=z=1$
