Chủ đề này có 2 trả lời

[DOTOANNANG]

Nếu $a,\,b,\,c,\,d > 0$ thỏa $ac+ bd= 2$ thì:

 

$$\frac{16}{a^{2}+ b^{2}+ c^{2}+ d^{2}}\leqq \sum\limits_{sym}\frac{1}{a}$$

 

$$\sum\limits_{cyc}\frac{a\left ( b+ c \right )}{a^{2}+ bc} \leqq \sum\limits_{sym}\frac{1}{a}$$

 

$$\frac{16}{a^{2}+ b^{2}+ c^{2}+ d^{2}} \leqq \sum\limits_{cyc}\frac{a\left ( b+ c \right )}{a^{2}+ bc} \,?$$

 

 

[Arthur Pendragon]

Con bất đầu tiên:

$\frac{16}{a^2+b^2+c^2+d^2} \leq \frac{16}{2ac+2bd} = \frac{16}{4}=4 (1)$

$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\geq \left(\dfrac{2}{\sqrt{ac}}+\dfrac{2}{\sqrt{bd}}\right)^2 \geq \dfrac{64}{(\sqrt{ac}+\sqrt{bd})^2} \geq \dfrac{64}{2(ac+bd)}=\dfrac{64}{4}=16$

$\Rightarrow \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d} \geq 4$ (2)

 

Từ (1) và (2) suy ra đpcm.

 

[DOTOANNANG]

Nếu $a,\,b,\,c,\,d > 0$ thỏa $ac+ bd= 2$ thì:

 

$$\frac{16}{a^{2}+ b^{2}+ c^{2}+ d^{2}}\leqq \sum\limits_{sym}\frac{1}{a}$$

 

$$\sum\limits_{cyc}\frac{a\left ( b+ c \right )}{a^{2}+ bc} \leqq \sum\limits_{sym}\frac{1}{a}$$

 

$$\frac{16}{a^{2}+ b^{2}+ c^{2}+ d^{2}} \leqq \sum\limits_{cyc}\frac{a\left ( b+ c \right )}{a^{2}+ bc} \,?$$

 

[$1$. kết quả không mạnh]

 

$$\frac{16}{\sum\limits_{sym}a^{2} }\leqq \sum\limits_{cyc}\frac{a\left ( b+ c \right )}{a^{2}+ bc}\leqq \sum\limits_{sym}\frac{1}{a}$$