Đáp án:
Giải thích các bước giải:
$z \geq x+y ⇒\dfrac{z}{x+y} \geq 1$
Đặt vế trái của BĐT là P
Ta có: $x^2+y^2 \geq \dfrac{1}{2}(x+y)^2$
$\dfrac{1}{x^2}+\dfrac{1}{y^2} \geq \dfrac{1}{2}\left( \dfrac{1}{x}+\dfrac{1}{y}\right)^2 \geq \dfrac{1}{2}\left( \dfrac{4}{x+y} \right)^2=\dfrac{8}{(x+y)^2}$
Do đó:
$P \geq \left( \dfrac{1}{2}(x+y)^2+z^2\right)\left(\dfrac{8}{(x+y)^2}+\dfrac{1}{z^2} \right)$
$⇒P \geq 5+\dfrac{1}{2}\left(\dfrac{x+y}{z} \right)^2+8\left(\dfrac{z}{x+y} \right)^2$
$⇒P \geq 5+\dfrac{1}{2}\left(\dfrac{x+y}{z} \right)^2+\dfrac{1}{2}\left(\dfrac{z}{x+y} \right)^2+\dfrac{15}{2}\left(\dfrac{z}{x+y} \right)^2$
$⇒P \geq 5+\dfrac{1}{2}.2\sqrt{\left(\dfrac{x+y}{z} \right)^2·\left(\dfrac{z}{x+y} \right)^2}+\dfrac{15}{2}·1=\dfrac{27}{2}$ (đpcm)
Dấu "=" xảy ra khi $2x=2y=z$
