Cho các số thực dương $a,b,c$ thỏa mãn: $a^2+2b^2+3c^2=1$. Tìm giá trị nhỏ nhất của biểu thức: $P=2a^3+3b^3+4c^3$
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$2P=2(a^3+a^3+x^3)+3(b^3+b^3+y^3)+4(c^3+c^3+z^3)-(2x^3+3y^3+4z^3) \geq 6xa^2+9yb^2+12zc^2-(2x^3+3y^3+4z^3)$
chọn $2x=\frac{3}{2}y=\frac{4}{3}z ,a=x,b=y,c=z, a^2+2b^2+3c^2=1$
$=> x=\frac{6}{\sqrt{407}}$ =>......
sao kq nó lẽ vậy nhỉ
Dùng bdt holder ta có 
$P^2.\frac{407}{144}=(2a^3+3b^3+4c^3)(2a^3+3b^3+4c^3)(\frac 14+\frac 89+\frac{27}{16}) \ge (a^2+2b^2+3c^2)^3=1$
$\Leftrightarrow P \ge \frac{12}{\sqrt{407}}$
Đẳng thức xảy ra $\Leftrightarrow a=\frac{6}{\sqrt{407}},b=\frac{8}{\sqrt{407}},c=\frac{9}{\sqrt{407}}$

ok có hơi nhầm tí –  tran85295 23-05-16 10:38 PM
kia là 2a^3, 3b^3,4c^3 mà Nam –  ๖ۣۜPXM๖ۣۜMinh4212♓ 23-05-16 10:06 PM

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