1,cho $0<a<1$ c/m:
$a(1-a^2) \leqslant \frac{2}{3*\sqrt{3}}$
2,cho a$> 0$$> 0$
c/m $a^2.(1-2a) \leqslant.\frac{1}{27}$
3,cho $a,b,c > 0$ t/m $a+b+c=3$
c/m:$a+\sqrt{ab}+3.\sqrt{abc}\leqslant 4$
Edited by nguyentrunghieua, 20-07-2013 - 08:18.
1,cho $0<a<1$ c/m:
$a(1-a^2) \leqslant \frac{2}{3*\sqrt{3}}$
2,cho a$> 0$$> 0$
c/m $a^2.(1-2a) \leqslant.\frac{1}{27}$
3,cho $a,b,c > 0$ t/m $a+b+c=3$
c/m:$a+\sqrt{ab}+3.\sqrt{abc}\leqslant 4$
Edited by nguyentrunghieua, 20-07-2013 - 08:18.
ONE PIECE IS THE BEST
câu 2 $a.a(1-2a)\leq \frac{(a+a+1-2a)^{3}}{27}=\frac{1}{27}$
tàn lụi
1,cho $0<a<1$ c/m:
$a(1-a^2) \leqslant \frac{2}{3*\sqrt{3}}$
2,cho a$> 0$$> 0$
c/m $a^2.(1-2a) \leqslant.\frac{1}{27}$
3,cho $a,b,c > 0$ t/m $a+b+c=3$
c/m:$a+\sqrt{ab}+3.\sqrt{abc}\leqslant 4$
câu 3 thay a=b=c=1 thì BĐT không đúng
3,cho $a,b,c > 0$ t/m $a+b+c=3$
c/m:$a+\sqrt{ab}+3.\sqrt{abc}\leqslant 4$
chắc đề phải là: $a+\sqrt{ab}+\sqrt[3]{abc} \leq 4$ chứ nhỉ!
Ta có:
$2\sqrt{ab} \leq \frac{a}{2}+2b$
$\Rightarrow$ $\sqrt{ab} \leq \frac{a}{4}+b$
$3\sqrt[3]{abc} \leq \frac{a}{4}+b+4c$
$\Rightarrow$ $\sqrt[3]{abc} \leq \frac{a}{12}+\frac{b}{3}+\frac{4c}{3}$
Cộng vào ta có đpcm
Edited by trandaiduongbg, 20-07-2013 - 14:29.
1,cho $0<a<1$ c/m:
$a(1-a^2) \leqslant \frac{2}{3*\sqrt{3}}$
ta có a$a^{3}+\frac{1}{3\sqrt{3}}+\frac{1}{3\sqrt{3}}\geq 3\sqrt[3]{\frac{a^{3}}{(3\sqrt{3})^{2}}}= a\Rightarrow \frac{2}{3\sqrt{3}}\geq a(1-a^{2})$
1,cho $0<a<1$ c/m:
$a(1-a^2) \leqslant \frac{2}{3*\sqrt{3}}$
2,cho a$> 0$$> 0$
c/m $a^2.(1-2a) \leqslant.\frac{1}{27}$
3,cho $a,b,c > 0$ t/m $a+b+c=3$
c/m:$a+\sqrt{ab}+3.\sqrt{abc}\leqslant 4$ (chỗ này là $\sqrt[3]{abc}$ chứ không phải là $3\sqrt[3]{abc}$)
ZION
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