7 replies to this topic

[ducpham]

Cho a, b, c > 0 và $ab+bc+ca\geq 3$ chứng minh $\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq\frac{3}{2}$

Edited by hachinh2013, 21-02-2015 - 23:03.

[vda2000]

cho a,b,c>0 và $ab+bc+ca\geq 3$ chứng minh $\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq\frac{3}{2}$

 Ta có:

$\sum\frac{a^3}{b+c}=\sum\frac{a^4}{ab+ac}$.

Áp dụng bất đẳng thức $Cauchy-Schwarz$, ta có:

$\sum\frac{a^4}{ab+ac}\geq\frac{(a^2+b^2+c^2)^2}{ab+ac+ab+bc+ca+bc}=\frac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}$

 

Cần chứng minh: $\frac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\geq\frac{3}{2}$

                      <=> $(a^2+b^2+c^2)^2\geq 3(ab+bc+ca) (1)$

 

Áp dụng bất đẳng thức phụ: $a^2+b^2+c^2\geq ab+bc+ca$

=> $(a^2+b^2+c^2)\geq (ab+bc+ca)^2\geq 3(ab+bc+ca)$ vì theo gt: $ab+bc+ca\geq 3$

=> $(1)$ đúng.

=>$Q.E.D$

Edited by vda2000, 21-02-2015 - 23:07.

[Ngoc Hung]

Cho a, b, c > 0 và $ab+bc+ca\geq 3$ chứng minh $\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq\frac{3}{2}$

Ta có $\frac{a^{3}}{b+c}+\frac{b^{3}}{c+a}+\frac{c^{3}}{a+b}=\frac{a^{4}}{ab+ca}+\frac{b^{4}}{bc+ab}+\frac{c^{4}}{ca+bc}\geq \frac{(a^{2}+b^{2}+c^{2})^{2}}{2(ab+bc+ca)}\geq \frac{a^{2}+b^{2}+c^{2}}{2}\geq \frac{ab+bc+ca}{2}\geq \frac{3}{2}$

[Namthemaster1234]

$\frac{a^3}{b+c}+\frac{b+c}{4}+\frac{1}{2}\geqslant 3\sqrt[3]{\frac{a^3.(b+c)}{8(b+c)}}=\frac{3a}{2}$

 

$\rightarrow \sum \frac{a^3}{b+c}+\frac{a+b+c}{2}+\frac{3}{2}\geqslant \frac{3(a+b+c)}{2}\rightarrow \sum \frac{a^3}{b+c}\geq (a+b+c)-\frac{3}{2}$

 

Để ý rằng : $(a+b+c)^2\geq 3(ab+ac+bc)=9\rightarrow a+b+c\geqslant 3$

 

Từ đó suy ra đpcm

[huykinhcan99]

Theo bất đẳng thức $\rm{H\ddot{o}lder}$ ta có:

\begin{align} & \left(\sum_{cyc} a\right)^3=\left(\sum_{cyc} 1.\dfrac{a}{\sqrt[3]{b+c}}.\sqrt[3]{b+c}\right)^3\leqslant \left(1+1+1\right) \left( \sum_{cyc} \dfrac{a^3}{b+c}\right)\left[\sum_{cyc} \left( b+c\right) \right] \nonumber \\ \Leftrightarrow & (a+b+c)^3\leqslant 6\left(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\right)(a+b+c) \nonumber \\ \label{eq:1} \Leftrightarrow & \ \dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\geqslant \dfrac{\left(a+b+c\right)^2}{6} \end{align}

 

Mặt khác, theo bất đẳng thức $\rm{AM - GM}$ thì ta có:

\begin{align} \label{eq:2} & \dfrac{a^2+b^2}{2} \geqslant ab \\ \label{eq:3} & \dfrac{b^2+c^2}{2} \geqslant bc \\ \label{eq:4} & \dfrac{c^2+a^2}{2} \geqslant ca \end{align}

 

Cộng các bất đẳng thức \eqref{eq:2}, \eqref{eq:3}, \eqref{eq:4} vế với vế ta thu được:

\begin{align} a^2+b^2+c^2 & \geqslant ab+bc+ca \nonumber \\ \label{eq:5} \Rightarrow (a+b+c)^2 & \geqslant 3(ab+bc+ca) \end{align}

 

Theo giả thiết ta lại có:

\begin{equation} \label{eq:6} ab+bc+ca \geqslant 3 \end{equation}

 

Từ các bất đẳng thức \eqref{eq:1}, \eqref{eq:5}, \eqref{eq:6} ta thu được 

\begin{equation} \dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\geqslant \dfrac{\left(a+b+c\right)^2}{6} \geqslant \dfrac{3(ab+bc+ca)}{6}\geqslant \dfrac{3}{2} \tag{$\blacksquare$} \end{equation}

 

Dấu bằng xảy ra khi $a=b=c$

[Coppy dera]

Đó là cách lớp mấy z

[dera coppy]

không hiểu

[huykinhcan99]

Bất đẳng thức Holder: (với mọi số thực $a_{1_1}$, $a_{1_2}$, $\ldots$, $a_{n_m}$)

$$\left(\sum_{i=1}^{n}a_{i_1}^m\right) \left(\sum_{i=1}^{n}a_{i_2}^m\right) \ldots \left(\sum_{i=1}^{n}a_{i_m}^m\right)\geqslant \left(\sum^n_{i=1}a_{i_1} a_{i_2} \ldots a_{i_m} \right)^m$$

hay có thể được viết

\begin{multline*} \left(a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m\right) \left(a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m\right) \ldots \left(a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m\right)\geqslant \\ \geqslant \left(a_{1_1} a_{1_2} \ldots a_{1_m}+a_{2_1} a_{2_2} \ldots a_{2_m} + \ldots+ a_{n_1} a_{n_2} \ldots a_{n_m} \right)^m \end{multline*}

 

Chứng minh:

Theo bất đẳng thức $AM - GM$, ta có: 

\begin{multline*} \dfrac{a_{1_1}^m}{a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m}+\dfrac{a_{1_2}^m}{a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m} + \ldots + \dfrac{a_{1_m}^m}{a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m}\geqslant \\ \geqslant m\dfrac{a_{1_1} a_{1_2} \ldots a_{1_m}}{\sqrt[m]{\left(a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m\right) \left(a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m\right) \ldots \left(a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m\right)}} \end{multline*}

 

Tương tự:

\begin{multline*} \dfrac{a_{2_1}^m}{a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m}+\dfrac{a_{2_2}^m}{a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m} + \ldots + \dfrac{a_{2_m}^m}{a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m}\geqslant \\ \geqslant m\dfrac{a_{2_1} a_{2_2} \ldots a_{2_m}}{\sqrt[m]{\left(a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m\right) \left(a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m\right) \ldots \left(a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m\right)}} \end{multline*}

$$\ldots$$

\begin{multline*} \dfrac{a_{n_1}^m}{a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m}+\dfrac{a_{n_2}^m}{a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m} + \ldots + \dfrac{a_{n_m}^m}{a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m}\geqslant \\ \geqslant m\dfrac{a_{n_1} a_{n_2} \ldots a_{n_m}}{\sqrt[m]{\left(a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m\right) \left(a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m\right) \ldots \left(a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m\right)}} \end{multline*}

 

Cộng $n$ bất đẳng thức trên lại ta có:

$$m\geqslant m\dfrac{a_{1_1} a_{1_2} \ldots a_{1_m}+a_{2_1} a_{2_2} \ldots a_{2_m}+\ldots+a_{n_1} a_{n_2} \ldots a_{n_m}}{\sqrt[m]{\left(a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m\right) \left(a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m\right) \ldots \left(a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m\right)}}$$

 

hay là

\begin{multline*} \sqrt[m]{\left(a_{1_1}^m+a_{2_1}^m+\ldots+a_{n_1}^m\right) \left(a_{1_2}^m+a_{2_2}^m+\ldots+a_{n_2}^m\right) \ldots \left(a_{1_m}^m+a_{2_m}^m+\ldots+a_{n_m}^m\right)} \geqslant \\ \geqslant a_{1_1} a_{1_2} \ldots a_{1_m}+a_{2_1} a_{2_2} \ldots a_{2_m}+\ldots+a_{n_1} a_{n_2} \ldots a_{n_m} \end{multline*}

 

Từ đây ta lấy lũy thừa bậc $m$ nhận được ngay bất đẳng thức cần chứng minh