Bđt(đại số 9)

C

congchuaanhsang

Đặt x=$a^3$ ; y=$b^3$ ; z=$c^3$\Rightarrowa,b,c\geq0
\RightarrowVT=$4a^3b^3$+$4b^3c^3$+$4c^3a^3$ ; VP=$4c^6$+$(a^2+b^2)^3$
VP - VT=$4c^6$+$a^6$+$3a^4b^2$+$3a^2b^4$+$b^6$-$4a^3b^3$-$4b^3c^3$-$4c^3a^3$
=($a^6$+2$a^3b^3$+$b^6$)-$4c^3$($a^3$+$b^3$)+$4c^6$+3($a^4b^2$-$2a^3b^3$+$a^2b^4$)
=$(a^3+b^3)^2$-$4c^3$($a^3$+$b^3$)+$4c^6$+3$(a^2b-ab^2)^2$
=$(a^3+b^3-2c^3)^2$+$3(a^2b-ab^2)^2$\geq0
\RightarrowVP\geqVT
\Leftrightarrow$4(\sqrt[3]{x^3y^3}+\sqrt[3]{y^3z^3}+\sqrt[3]{z^3x^3})$\leq$4z^3$+$(x+y)^3$
 
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