Chủ đề này có 2 trả lời

[lymiu]

 Cm bất đẳng thức sau :

      $\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{c}+\frac{1}{d}}\leqslant \frac{1}{\frac{1}{a+c}+\frac{1}{b+d}}$

[vipboycodon]

mình nhầm ạ!!!

Bài viết đã được chỉnh sửa nội dung bởi vipboycodon: 10-01-2015 - 22:11

[huykinhcan99]

 Cm bất đẳng thức sau :

      $\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{c}+\frac{1}{d}}\leqslant \frac{1}{\frac{1}{a+c}+\frac{1}{b+d}}$

 

Theo bất đẳng thức $Cauchy$ ta có:

$$(a+c)^2=\left(\dfrac{a}{\sqrt{a+b}}.\sqrt{a+b}+\dfrac{c}{\sqrt{c+d}}.\sqrt{c+d}\right)^2\leqslant \left(\dfrac{a^2}{a+b}+\dfrac{c^2}{c+d}\right)\left(a+b+c+d\right)$$

 

Từ đó ta suy ra \begin{equation} \label{eq:1} \tag{1} \dfrac{a^2}{a+b}+\dfrac{c^2}{c+d} \geqslant \dfrac{(a+c)^2}{a+b+c+d}\end{equation}

 

Biến đổi tương đương ta thu được

\begin{align} \eqref{eq:1} \ & \Leftrightarrow \dfrac{a^2+ab-ab}{a+b}+\dfrac{c^2+cd-cd}{c+d} \geqslant \dfrac{(a+c)^2+(a+c)(b+d)-(a+c)(b+d)}{a+b+c+d} \nonumber \\ & \Leftrightarrow \left(a-\dfrac{ab}{a+b}\right)+\left(c-\dfrac{cd}{c+d}\right) \geqslant (a+c)-\dfrac{(a+c)(b+d)}{a+b+c+d} \nonumber \\ & \Leftrightarrow \left(\dfrac{ab}{a+b}-a\right)+\left(\dfrac{cd}{c+d}-c\right)\leqslant \dfrac{(a+c)(b+d)}{(a+c)+(b+d)}-(a+c) \nonumber \\ &\Leftrightarrow \dfrac{ab}{a+b}+\dfrac{cd}{c+d}\leqslant \dfrac{(a+c)(b+d)}{(a+c)+(b+d)} \nonumber \\ &\Leftrightarrow \dfrac{1}{\dfrac{a+b}{ab}}+\dfrac{1}{\dfrac{c+d}{cd}}\leqslant \dfrac{1}{\dfrac{(a+c)+(b+d)}{(a+c)(b+d)}} \nonumber \\ \tag{$\blacksquare$} &\Leftrightarrow \dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{d}}\leqslant \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}} \end{align}

 

Dấu bằng xảy ra khi $\dfrac{a}{b}=\dfrac{c}{d}$