$x,y,z>0,x+y+z=\frac{3}4$

tìm $GTLN$ $P=\sqrt[3]{x+3y}+\sqrt[3]{y+3z}+\sqrt[3]{z+3x}$


post mn làm chứ k hỏi nhá.
co dua cho tap tai lieu day cop toan bdt k ak. hay cuc lun. tai lieu cua chuyen Tran Phu HP –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 16-04-14 01:09 PM
cosi cho 3 so thi de hon ak. k nham thi trong sach co hai cach lam hay sao ak. –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 16-04-14 01:08 PM
BĐT holder là ra :P –  ♂Vitamin_Tờ♫ 16-04-14 08:56 AM
Tui hk bít gõ latex mong bác thông kảm nghen
Viết lại biểu thức thành P=cănbậc3[1.1.(x+3y)] +cănbậc3[1.1.(y+3z)]+cănbậc3[1.1.(z+3x)]
áp dụng BĐT AM-GM ta có P=< (1+1+x+3y)/3 +(1+1+y+3z)/3 +(1+1+z+3x)/3 
=>P=<[4(x+y+z)+6]/3 =3
Vậy Pmax =3 xảy ra khi x=y=z=1/4
p/s:bít gõ latex thì chỉ tui với :3
xem o huong dan go cong thuc nhe. cho vao giua $$ thi se hien cong thuc. –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 17-04-14 07:32 PM
http://toan.hoctainha.vn/faq –  ♥♥♥ Panda Sơkiu Panda Mập ♥♥♥ 17-04-14 07:31 PM

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