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Question

1) Cho $0\leq x,y,z\leq1$. Chứng minh:

$2(x^3+y^3+z^3)-(x^2y+y^2z+z^2x)\leq3$

2) Cho $a,b,c >0$ và $a+b+c=1$. Tìm min:

M=$\frac{a^3}{1-a^2}+\frac{b^3}{1-b^2}+\frac{c^3}{1-c^2}$

3) Cho $a,b,c >0$ và $a+b+c\geq 6$. Tìm min:

M=$\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}$

khangnguyenthanh · Answer 1 · 9/1/2013 4:52:17 PM

Câu 3:

Áp dụng BĐT Mincopxki ta có:

$M\geq \sqrt{(a+b+c)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}$

$\geq\sqrt{(a+b+c)^2+\dfrac{81}{(a+b+c)^2}}$

$=\sqrt{(a+b+c)^2+\dfrac{1296}{(a+b+c)^2}-\dfrac{1215}{(a+b+c)^2}}$

$\geq\sqrt{2\sqrt{(a+b+c)^2.\dfrac{1296}{(a+b+c)^2}}-\dfrac{1215}{36}}=\dfrac{3\sqrt{17}}{2}$

$\min M=\dfrac{3\sqrt{17}}{2} \Leftrightarrow a=b=c=2$

Sửa 01-09-13 04:52 PM

khangnguyenthanh
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Trả lời 01-09-13 04:51 PM

khangnguyenthanh
72K 2 480 42

332K 4M

1

khangnguyenthanh · Answer 2 · 8/31/2013 9:08:32 PM

1. Ta có:

$(1-y)(1-x^{2})\geq 0\Rightarrow 1-y-x^{2}+2x^{3}\geq 2x^{3}-x^{2}y$

Tương tự $1-z-y^{2}+2y^{3}\geq 2y^{3}-y^{2}z,1-x-z^{2}+2z^{3}\geq 2z^{3}-z^{2}x$

$\Rightarrow 2(x^{3}+y^{3}+z^{3})-(x^{2}y+y^{2}z+z^{2}x)\leq 3-x-y-z-x^{2}-y^{2}-z^{2}+2(x^{3}+y^{3}+z^{3})\leq 3$

do $x^{3}\leq x^{2},x^{3}\leq x,y^{3}\leq y^{2},y^{3}\leq y,z^{3}\leq z^{2},z^{3}\leq z$

Trả lời 31-08-13 09:08 PM

khangnguyenthanh
72K 2 480 42

332K 4M

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