Toán Lớp 9: cho a,b,c >0 và a+b+c=1
CMR: a/1+a+b/1+b+c/1+c <=3/4
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Toán Lớp 9: cho a,b,c >0 và a+b+c=1
CMR: a/1+a+b/1+b+c/1+c <=3/4
Comments ( 2 )
Vì nếu tách ra thì = a+b+c+a/1+b/1+c/1
>1 vì a+b+c = 1, a/1>0,b/1>0,c/1>0 ( a,b,c>0)
=> a/1+a+b/1+b+c/1+c>1>3/4
$$\dfrac{a^2}{m}+\dfrac{b^2}{n}+\dfrac{c^2}{p} \ge \dfrac{(a+b+c)^2}{m+n+p} $$
Dấu đẳng thức xảy ra khi$$\dfrac{a}{m}=\dfrac{b}{n}=\dfrac{c}{p}$$
$$\begin{array}{lrl}
&\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}&\le\dfrac{3}{4}\\
\Leftrightarrow&\left( 1-\dfrac{1}{1+a} \right)+\left( 1-\dfrac{1}{1+b} \right)+\left( 1-\dfrac{1}{1+c} \right)\le \dfrac{3}{4}\\
\Leftrightarrow&\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge \dfrac{9}{4}
\end{array}$$
Thật vậy, áp dụng bổ đề ta có:
$$\begin{array}{rl}
&\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\\
=&\dfrac{1^2}{1+a}+\dfrac{1^2}{1+b}+\dfrac{1^2}{1+c}\\
\ge&\dfrac{(1+1+1)^2}{3+a+b+c}=\dfrac{3^2}{3+1}=\dfrac{9}{4}(\text{đpcm})
\end{array}$$