Bài 104324

Question

Cho $n \in Z,n \geq 1; a,b,c>0$.Chứng minh rằng:
$\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geq \frac{3}{2}(\frac{a+b+c}{3})^{n-1}$

huyenhana164 · Answer 1 · 5/15/2012 6:29:59 PM

Xét $n=1:$
BĐT trở thành : $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$

$VT=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\geq\frac{(a+b+c)^2}{2(ab+bc+ca)} $( áp dụng BĐT Bunhiacopxki)

Mà $ ab+bc+ca\leq \frac{(a+b+c)^2}{3}$

$\Rightarrow VT\geq\frac{3}{2}$ (dpcm)

Xét $n \geq 2$
Xét $3$ dãy số:
$\frac{a}{\sqrt[n]{b+c}},\sqrt[n]{b+c},\underbrace {1,1,...,1}_{n-2 số};$
$\frac{b}{\sqrt[n]{c+a}},\sqrt[n]{c+a},\underbrace {1,1,...,1}_{n-2 số};$
$\frac{c}{\sqrt[n]{a+b}},\sqrt[n]{a+b},\underbrace {1,1,...,1}_{n-2 số}$
Áp dụng BĐT Bunhiacopski(mở rộng):
$(\frac{a}{\sqrt[n]{b+c}}.\sqrt[n]{b+c}.1.1..1+\frac{b}{\sqrt[n]{c+a}}.\sqrt[n]{c+a}.1.1..1+\frac{c}{\sqrt[n]{a+b}}.\sqrt[n]{a+b}.1.1..1)$
$\leq (\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b})(b+c+c+a+a+b).\underbrace {(1+1+1)+...+(1+1+1)}_{n-2 thừa số}$
$\Rightarrow (a+b+c)^{n} \leq (\frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b})2(a+b+c).3^{n-2}$
$\Rightarrow \frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geq \frac{3}{2}(\frac{a+b+c}{3})^{n-1}$
Dấu "=" xảy ra $\Leftrightarrow a=b=c$

Bài 104324

1 Đáp án

Lý thuyết liên quan

Bài 104324

1 Đáp án

Liên quan

Lý thuyết liên quan