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(jiàn)解析 (2) 18

【解析】(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)?/span>5x+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.