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a;b;c>0. Chứng minh $\frac{a^{3}}{(b+c)^{2}} + \frac{b^{3}}{(c+a)^{2}} + \frac{c^{3}}{(a+b)^{2}} \geq \frac{a+b+c}{2}$

khangnguyenthanh · Answer 1 · 9/30/2014 10:41:10 AM

Áp dụng BĐT Cauchy ta có:

$\dfrac{a^3}{(b+c)^2}+\dfrac{b+c}{8}+\dfrac{b+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{(b+c)^2}.\dfrac{b+c}{8}.\dfrac{b+c}{8}}=\dfrac{3a}{4}$

$\dfrac{b^3}{(c+a)^2}+\dfrac{c+a}{8}+\dfrac{c+a}{8}\ge3\sqrt[3]{\dfrac{b^3}{(c+a)^2}.\dfrac{c+a}{8}.\dfrac{c+a}{8}}=\dfrac{3b}{4}$

$\dfrac{c^3}{(a+b)^2}+\dfrac{a+b}{8}+\dfrac{a+b}{8}\ge3\sqrt[3]{\dfrac{c^3}{(a+b)^2}.\dfrac{a+b}{8}.\dfrac{a+b}{8}}=\dfrac{3c}{4}$

Suy ra: $\dfrac{a^3}{(b+c)^2}+\dfrac{b^3}{(c+a)^2}+\dfrac{c^3}{(a+b)^2}\ge\dfrac{a+b+c}{4}$

Dấu bằng xảy ra khi $a=b=c$.

Trả lời 30-09-14 10:41 AM

khangnguyenthanh
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WhjteShadow · Answer 2 · 9/28/2014 9:57:32 PM

Đề bài như sau:Cho a,b,c >0 CM:$\frac{a^3}{b(a+c)}+\frac{b^3}{c(a+b)}+\frac{c^3}{a(b+c)}\geq \frac{a+b+c}{2}$

Có $\frac{a^3}{b(a+c)}+\frac{b}{2}+\frac{a+c}{4}\geq \frac{3}{2}a$

Tương tự $\frac{b^3}{c(a+b)}+\frac{c}{2}+\frac{a+b}{4}\geq \frac{3}{2}.b;\frac{c^3}{a(b+c)}+\frac{a}{2}+\frac{b+c}{4}\geq \frac{3}{2}c$

$\Rightarrow VT\geq \frac{3}{2}(a+b+c)-(\frac{a+b+c}{2}+\frac{2(a+b+c)}{4})=\frac{a+b+c}{2}$(đpcm)

Dấu = xảy ra khi a=b=c.

Trả lời 28-09-14 09:57 PM

WhjteShadow
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