Lời giải
Đề bài:
Cho $a,b,c$ là $3$ cạnh $\triangle ABC$.Chứng minh rằng:$a) \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \geq 3$$b) \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \sqrt{\frac{3}{2Rr}}$(Với $R,r$ là bán kính đường tròn ngoại,nội tiếp $\triangle ABC$ tương ứng)
Lời giải
a)Đặt: $\begin{cases}x=b+c-a >0\\ y=c+a-b>0 \\z=a+b-c>0 \end{cases}$
$\Rightarrow \begin{cases}a=\frac{y+z}{2} \\ b=\frac{z+x}{2} \\c=\frac{x+y}{2}\end{cases}$
Suy ra:
Vế trái BĐT=$\frac{y+z}{2x}+\frac{z+x}{2y}+\frac{x+y}{2z}=\frac{1}{2}(\frac{x}{y}+\frac{y}{x})+\frac{1}{2}(\frac{y}{z}+\frac{z}{y})+\frac{1}{2}(\frac{z}{x}+\frac{x}{z})$
$\geq 1+1+1=3$
$\Rightarrow $ (ĐPCM)
Dấu “=” xảy
ra khi $a=b=c$
b)Ta có:$S=\frac{abc}{4R}=pr\Rightarrow 2Rr=\frac{abc}{a+b+c}$
$\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt{\frac{3}{2Rr}}\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \sqrt{\frac{3(a+b+c)}{abc}}$
$\Rightarrow (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}\geq 3(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}) (1)$
Theo BĐT Cauchy:
$ (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})=$
$=\frac{1}{2}(\frac{1}{a^{2}}+\frac{1}{b^{2}})+\frac{1}{2}(\frac{1}{b^{2}}+\frac{1}{c^{2}})+\frac{1}{2}(\frac{1}{c^{2}}+\frac{1}{a^{2}})+2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$
$\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}++2(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})$
$=3(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\Rightarrow (1)$đúng
$\Rightarrow $ (ĐPCM)
Dấu “=” xảy
ra khi $a=b=c$
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Chuyên mục: Bất đẳng thức Côsi
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