Cho a, b, c là ba cạnh tam giác. Chứng minh:
a) $\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}\geq 1$
b) $\frac{a}{\sqrt{b+c-a}}+\frac{b}{\sqrt{a+c-b}}+\frac{c}{\sqrt{a+b-c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}$
Cho a,b,c là ba cạnh t.giác.C/m$\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}\geq 1$
Started By nhox sock tn, 06-09-2013 - 17:31
#1
Posted 06-09-2013 - 17:31
nhox sock tn
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#2
Posted 06-09-2013 - 17:48
Zaraki
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PQT
Cho a, b, c là ba cạnh tam giác. Chứng minh:
a) $\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}\geq 1$
Lời giải. Đặt $\begin{cases} 2b+2c-a=x \\ 2c+2a-b=y \\ 2a+2b-c=z \end{cases}$. Khi đó ta có $a= \frac{2y+2z-x}{9},b= \frac{2x+2z-y}{9}, c= \frac{2x+2y-z}{9}$. Bất đẳng thức cần chứng minh tương đương với $$\frac 29 \left[ \left( \frac yx+ \frac xy \right)+ \left( \frac zx+ \frac xz \right)+ \left( \frac yz + \frac zy \right) \right] \ge \frac 43$$
Điều này hiển nhiên đúng theo BĐT AM-GM.
Dấu đẳng thức xảy ra khi và chỉ khi $a=b=c$.
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