Chủ đề này có 1 trả lời

[Xuan Hung HQH]

Cho a,b,c dương TM:$ab+bc+ac\geq 1$.CMR:$\frac{1}{\sqrt{a^2+ab+b^2}}+\frac{1}{\sqrt{b^2+bc+c^2}}+\frac{1}{\sqrt{c^2+ca+a^2}}\geq \frac{9}{(a+b+c)^2}$

Bài viết đã được chỉnh sửa nội dung bởi Xuan Hung HQH: 16-11-2014 - 13:32

[huykinhcan99]

Cho a,b,c dương TM:$ab+bc+ac\geq 1$.CMR:$\frac{1}{\sqrt{a^2+ab+b^2}}+\frac{1}{\sqrt{b^2+bc+c^2}}+\frac{1}{\sqrt{c^2+ca+a^2}}\geq \frac{9}{(a+b+c)^2}$

 

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

$$9=(1+1+1)^2 = \left(\sqrt[4]{a^2+ab+b^2}.\tfrac{1}{\sqrt[4]{a^2+ab+b^2}}+\sqrt[4]{b^2+bc+c^2}.\tfrac{1}{\sqrt[4]{b^2+bc+c^2}}+\sqrt[4]{c^2+ca+a^2}.\tfrac{1}{\sqrt[4]{c^2+ca+a^2}}\right)^2 \leqslant \left(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\right) \left(\tfrac{1}{\sqrt{a^2+ab+b^2}}+\tfrac{1}{\sqrt{b^2+bc+c^2}}+\tfrac{1}{\sqrt{c^2+ca+a^2}}\right)$$

 

\begin{equation} \label{1} \tag{1} \small{\Rightarrow \dfrac{1}{\sqrt{a^2+ab+b^2}}+\dfrac{1}{\sqrt{b^2+bc+c^2}}+\dfrac{1}{\sqrt{c^2+ca+a^2}}\geqslant \dfrac{9}{\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}}} \end{equation}

 

Mặt khác, lại áp dụng bất đẳng thức $Cauchy-Schwarz$ ta có:

\begin{eqnarray} &\phantom{\Leftrightarrow}& \small{\left(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\right)^2} \leqslant (1+1+1) \small{\left(a^2+ab+b^2+b^2+bc+c^2+c^2+ca+a^2\right)} \nonumber \\ &\Leftrightarrow &\left(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\right)^2 \leqslant 3\left[2\left(a^2+b^2+c^2\right)+\left(ab+bc+ca\right)\right] \nonumber \\ &\Leftrightarrow &\left(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\right)^2 \leqslant 3\left[2\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right] \nonumber \\ &\Leftrightarrow &\left(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\right)^2 \leqslant 6\left(a+b+c\right)^2-9\left(ab+bc+ca\right) \nonumber \\ &\Leftrightarrow& \left(\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\right)^2 \leqslant 6\left(a+b+c\right)^2-9 \text{ (vì $ab+bc+ca\geqslant 1$)} \nonumber \\ \label{2} \tag{2} &\Leftrightarrow &\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2} \leqslant \sqrt{6\left(a+b+c\right)^2-9} \end{eqnarray}

 

Áp dụng bất đẳng thức $AM-GM$  cho hai số $(a+b+c)^4$ và $9$ ta có:

\begin{eqnarray} \ 2.\left(a+b+c\right)^2.3 \leqslant \left(a+b+c\right)^4+9 \nonumber \\ 6\left(a+b+c\right)^2-9 \leqslant \left(a+b+c\right)^4 \nonumber \\ \label{3} \tag{3} \sqrt{6\left(a+b+c\right)^2-9} \leqslant \left(a+b+c\right)^2  \end{eqnarray}

 

$$\eqref{1}, \eqref{2}, \eqref{3} \Rightarrow \dfrac{1}{\sqrt{a^2+ab+b^2}}+\dfrac{1}{\sqrt{b^2+bc+c^2}}+\dfrac{1}{\sqrt{c^2+ca+a^2}}\geqslant \dfrac{9}{(a+b+c)^2} \ _\square$$

 

Dấu bằng xảy ra khi $a=b=c=\dfrac{\sqrt{3}}{3}$