cho a,b,c >0 CM: a/ $\frac{a^{3}}{bc} + \frac{b^{3}}{ac}+\frac{c^{3}}{ab}\geq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$
b/cho $a^{2} +b^{2}+c^{2}=3$ , CM: $a^{3}+b^{3}+c^{3}\geq 3$
cho a,b,c >0 CM: a/ $\frac{a^{3}}{bc} + \frac{b^{3}}{ac}+\frac{c^{3}}{ab}\geq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$
b/cho $a^{2} +b^{2}+c^{2}=3$ , CM: $a^{3}+b^{3}+c^{3}\geq 3$
b/cho $a^{2} +b^{2}+c^{2}=3$ , CM: $a^{3}+b^{3}+c^{3}\geq 3$
Có: $9=3(a^2+b^2+c^2)\geq (a+b+c)^2\Rightarrow a+b+c\leq 3$. Do đó:
$3(a^3+b^3+c^3)\geq (a+b+c)(a^3+b^3+c^3)\geq (a^2+b^2+c^2)^2=9\Rightarrow a^3+b^3+c^3\geq 3( \text{đpcm})$
$\color{red}{\mathrm{\text{How I wish I could recollect, of circle roud}}}$
$\color{red}{\mathrm{\text{The exact relation Archimede unwound ! }}}$
cho a,b,c >0 CM: a/ $\frac{a^{3}}{bc} + \frac{b^{3}}{ac}+\frac{c^{3}}{ab}\geq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$
b/cho $a^{2} +b^{2}+c^{2}=3$ , CM: $a^{3}+b^{3}+c^{3}\geq 3$
b, Ta có: $3(a^2+b^2+c^2) \geq (a+b+c)^2 \iff a+b+c \leq \sqrt{3(a^2+b^2+c^2)}=3$
Ta có: $(a^3+b^3+c^3)(a+b+c) \geq (a^2+b^2+c^2)^2=9$
$\rightarrow a^3+b^3+c^3 \geq \dfrac{9}{a+b+c} \geq 3$
p/S: post chậm
Bài viết đã được chỉnh sửa nội dung bởi dunghoiten: 13-03-2016 - 11:03
cho a,b,c >0 CM: a/ $\frac{a^{3}}{bc} + \frac{b^{3}}{ac}+\frac{c^{3}}{ab}\geq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$
Ta có: $abc \leq \dfrac{(a^2+b^2+c^2)(a+b+c)}{9}$
Ta có: $\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4+b^4+c^4}{abc} \geq \dfrac{9(a^2+b^2+c^2)^2}{3(a+b+c)(a^2+b^2+c^2)}=\dfrac{3(a^2+b^2+c^2)}{a+b+c}$
Bài viết đã được chỉnh sửa nội dung bởi dunghoiten: 13-03-2016 - 11:08
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