(Toán 9) Tìm GTLN

H

hien_vuthithanh

Cho x,y,z>0 t/m: xyz=1
Tìm GTLN của bt:
$ Q= \dfrac{1}{x+y+1}+ \dfrac{1}{y+z+1}+\dfrac{1}{z+x+1}$
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Đặt: $x =a^3 ; y=b^3 ; z=c^3 \rightarrow abc=1$

$\rightarrow Q= \dfrac{1}{a^3+b^3+1}+ \dfrac{1}{b^3+c^3+1}+\dfrac{1}{c^3+a^3+1}$

$= \dfrac{1}{a^3+b^3+abc}+ \dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}$

$\le \dfrac{1}{ab(a+b)+abc}+ \dfrac{1}{bc(b+c)+abc}+\dfrac{1}{ca(a+c)+abc}$

$\le \dfrac{1}{a+b+c}.(\dfrac{1}{ab}+ \dfrac{1}{bc}+ \dfrac{1}{ca}) = \dfrac{1}{abc}=1$

$\rightarrow Q\le 1$

Dấu = tại $x=y=z=1$
 
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