Question

已知正數(shù)x,y,z滿足5x+4y+3z=10.
(1)求證:++≥5.
(2)求+的最小值.

Answer 1

(1)見解析   (2) 18

(1)根據(jù)柯西不等式,得
[(4y+3z)+(3z+5x)+(5x+4y)](++)≥(5x+4y+3z)²,
當(dāng)且僅當(dāng)==,
即x=,y=,z=時(shí)取等號(hào).
因?yàn)?x+4y+3z=10,
所以++≥=5.
(2)根據(jù)平均值不等式,得
+≥2=2·,
當(dāng)且僅當(dāng)x²=y²+z²時(shí),等號(hào)成立.
根據(jù)柯西不等式,得
(x²+y²+z²)(5²+4²+3²)≥(5x+4y+3z)²=100,
即x²+y²+z²≥2,當(dāng)且僅當(dāng)==時(shí),
等號(hào)成立.
綜上,+≥2·3²=18.
當(dāng)且僅當(dāng)x=1,y=,z=時(shí),等號(hào)成立.
所以+的最小值為18.