Cho a,b>0. CMR: $(\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}})\leq 2$
Cho a,b>0. CMR: $(\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}})\leq 2$
#1
Đã gửi 11-11-2018 - 08:02
#2
Đã gửi 11-11-2018 - 09:54
Cho a,b>0. CMR: $(\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}})\leq 2$
$P=(t+1)\frac{1}{\sqrt{1+3t^{2}}}+(t+1)\frac{1}{\sqrt{t^{2}+3}}; t=\sqrt{\frac{b}{a}}$;
Đạo hàm là xong
Bài viết đã được chỉnh sửa nội dung bởi anhtukhon1: 11-11-2018 - 09:54
#3
Đã gửi 11-11-2018 - 17:02
bạn có thể nói rõ hơn k, mình mới học lớp 9
#4
Đã gửi 11-11-2018 - 17:03
$P=(t+1)\frac{1}{\sqrt{1+3t^{2}}}+(t+1)\frac{1}{\sqrt{t^{2}+3}}; t=\sqrt{\frac{b}{a}}$;
Đạo hàm là xong
bạn có thể nói rõ hơn k, mình mới học lớp 9
#5
Đã gửi 11-11-2018 - 17:35
Nhờ các bạn trợ giúp bài toán
Cho x,y,z là các số thực dương thỏa mãn x $\geq$ z chứng minh rằng:
$\frac{xy}{y^{2}+yz}+\frac{y^{2}}{xz+yz}+\frac{x+2z}{x+z}\geqslant \frac{5}{2}$
#6
Đã gửi 14-11-2018 - 19:27
Đặt $\left\{\begin{matrix} a+ b= 2\,u\\ ab= v^{2} \end{matrix}\right.\,\rightarrow \,u\geqq v$
$$\left ( \sqrt{a}+ \sqrt{b} \right )\left ( \frac{1}{\sqrt{a+ 3\,b}}+ \frac{1}{\sqrt{b+ 3\,a}} \right )\leqq 2$$
$$\Leftrightarrow \left ( a+ b+ 2\,\sqrt{ab} \right )\left [ \frac{4\left ( a+ b \right )+ 2\,\sqrt{10\,ab+ 3\,a^{2}+ 3\,b^{2}}}{10\,ab+ 3\,a^{2}+ 3\,b^{2}} \right ]\leqq 4$$
$$\Leftrightarrow \left ( u+ v \right )\left ( \frac{\sqrt{3\,u^{2}+ v^{2}}+ 2\,u}{3\,u^{2}+ v^{2}} \right )\leqq 1$$
$$\Leftrightarrow \left ( 3\,u^{2}+ v^{2} \right )\left ( u+ v \right )^{2}\leqq 4\left ( 2\,u^{2}- uv+ v^{2} \right )$$
$$\Leftrightarrow \left ( u- v \right )\left [ 8\,v^{3}+ \left ( u- v \right )\left ( 13\,u^{2}+ 4\,uv+ 11\,v^{2} \right ) \right ]\geqq 0$$
#7
Đã gửi 16-11-2018 - 09:28
$$\left ( \sqrt{a}+ \sqrt{b} \right )\left ( \frac{1}{\sqrt{a+ 3\,b}}+ \frac{1}{\sqrt{b+ 3\,a}} \right )\leqq 2 \,\Leftrightarrow\,2\,\sqrt{\left ( a+ 3\,b \right )\left ( b+ 3\,a \right )} \geqq \left ( \sqrt{a}+ \sqrt{b} \right )\left ( \sqrt{3\,a+ b}+ \sqrt{3\,b+ a} \right )$$
$$2\,\frac{\sqrt{\left ( a+ 3\,b \right )\left ( b+ 3\,a \right )}}{\sqrt{a}+ \sqrt{b}} \geqq 2\,\sqrt{2\left ( a+ b \right )}\geqq \left ( \sqrt{3\,a+ b}+ \sqrt{3\,b+ a} \right )$$
#8
Đã gửi 16-11-2018 - 17:10
$$\left ( \sqrt{a}+ \sqrt{b} \right )\left ( \frac{1}{\sqrt{a+ 3\,b}}+ \frac{1}{\sqrt{b+ 3\,a}} \right )\leqq 2 \,\Leftrightarrow\,2\,\sqrt{\left ( a+ 3\,b \right )\left ( b+ 3\,a \right )} \geqq \left ( \sqrt{a}+ \sqrt{b} \right )\left ( \sqrt{3\,a+ b}+ \sqrt{3\,b+ a} \right )$$
$$2\,\frac{\sqrt{\left ( a+ 3\,b \right )\left ( b+ 3\,a \right )}}{\sqrt{a}+ \sqrt{b}} \geqq 2\,\sqrt{2\left ( a+ b \right )}\geqq \left ( \sqrt{3\,a+ b}+ \sqrt{3\,b+ a} \right )$$
$$\sqrt{\left ( a+ 3\,b \right )\left ( b+ 3\,a \right )}\geqq 3\left ( a+ b \right )+ 2\,\sqrt{ab}\geqq \sqrt{2\left ( a+ b \right )}\left ( \sqrt{a}+ \sqrt{b} \right )$$
#9
Đã gửi 16-11-2018 - 18:33
$$\sqrt{\left ( a+ 3\,b \right )\left ( b+ 3\,a \right )}\geqq_{\lfloor \,1\, \rceil} 3\left ( a+ b \right )+ 2\sqrt{ab}\geqq_{\lfloor \,2\, \rceil} \left ( \sqrt{a}+ \sqrt{b} \right )\left ( \sqrt{3\,a+ b}+ \sqrt{3\,b+ a} \right )$$
Đặt $\left\{\begin{matrix} a+ b= 2\,u\\ ab= v^{2} \end{matrix}\right.\,\rightarrow \,u\geqq v$
$\lfloor \,1\, \rceil\,\Leftrightarrow \,4\left ( 3\,u^{2}+ v^{2} \right )^{2}\geqq \left ( 3\,u+ v \right )^{2}\,\Leftrightarrow \,3\left ( u- v \right )^{2}\geqq 0$
$\lfloor \,2\, \rceil\,\Leftrightarrow \,\left ( 3\,u+ v \right )^{2}\geqq 2\left ( u+ v \right )\left ( 2\,u+ \sqrt{3\,u^{2}+ v^{2}} \right )\,$ $\Leftrightarrow \,\left [ \left ( 3\,u+ v \right )^{2}- 4\,u\left ( u+ v \right ) \right ]\geqq 4\left ( u+ v \right )^{2}\left ( 3\,u^{2}+ v^{2} \right )$ $\Leftrightarrow \left ( u- v \right )\left ( 13\,u^{3}+ 9\,u^{2}v+ 7\,uv^{2}+ 5\,v^{3} \right )\geqq 0$
#10
Đã gửi 16-11-2018 - 18:44
$$\sqrt{\left ( 3\,a+ b \right )\left ( 3\,b+ a \right )}\geqq \sqrt{a\left ( a+ 3\,b \right )}+ \sqrt{b\left ( b+ 3\,a \right )}+ \text{T}_{\lfloor \,3\, \rceil}$$
với:
$\text{T}_{\lfloor \,4\, \rceil}= \frac{5}{8}\left ( a+ b- 2\,\sqrt{ab} \right )\leqq 0$
#11
Đã gửi 16-11-2018 - 19:47
$-a+ 3\,b\,\,\geqq\, 0$
$$a+ 3\,b\,\,\geqq\, 2\,\cdot\,\frac{a}{a+ b}\,\,\sqrt{\,\,\frac{b\,\left ( -a+ 3\,b \right )}{2}\,\,}$$
#12
Đã gửi 21-11-2018 - 19:24
Đặt A=..... Ta cần cm A<=2
$A=\frac{\sqrt{a}}{\sqrt{a+3b}}+\frac{\sqrt{b}}{\sqrt{a+3b}}+\frac{\sqrt{a}}{\sqrt{b+3a}}+\frac{\sqrt{b}}{\sqrt{b+3a}}$
Với a,b>0 có:
$\frac{\sqrt{a}}{\sqrt{a+3b}}=\frac{\sqrt{a}}{\sqrt{a+b}}.\frac{\sqrt{a+b}}{\sqrt{a+3b}}\leq \frac{1}{2}(\frac{a}{a+b}+\frac{a+b}{a+3b})$
$\frac{\sqrt{b}}{\sqrt{a+3b}}=\frac{\sqrt{2b}}{\sqrt{a+3b}}.\frac{1}{\sqrt{2}}\leq \frac{1}{2}.(\frac{1}{2}+\frac{2b}{a+3b})$
Do đó $\frac{\sqrt{a}}{\sqrt{a+3b}}+\frac{\sqrt{b}}{\sqrt{a+3b}}\leq \frac{1}{2}(\frac{a}{a+b}+\frac{3}{2})$
Làm tương tự => đpcm
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