Chủ đề này có 6 trả lời

[PugMath]

Cho a,b,c > 0 và $(a+b+c)(ab+bc+ac)=9$

CMR : $3(a+b+c)\leqslant \sqrt[3]{26+a^3}+\sqrt[3]{26+b^3}+\sqrt[3]{26+c^3}$

--------------------By pugmath 

[DOTOANNANG]

Với $a,\,b,\,c> 0,\,\left ( a+ b+ c \right )\left ( ab+ bc+ ca \right )= 9$ thì:

 

 

 

 

$$\frac{3}{ab+ bc+ ca}\,\left ( ab^{2}+ bc^{2}+ ca^{2} \right )\geqq a+ b+ c\geqq a^{2}b^{2}+ b^{2}c^{2}+ c^{2}a^{2}$$

 

 

 

 

$$\sqrt{\frac{9}{abc}- \left ( 3- 2\sqrt{2} \right )\left ( 1- abc \right )}\geqq \sqrt{\frac{a+ b}{2\,c}}+ \sqrt{\frac{b+ c}{2\,a}}+ \sqrt{\frac{c+ a}{2\,b}}$$

(với $a,\,b,\,c$ là 3 cạnh của tam giác)

[DOTOANNANG]

$$3\left ( \frac{1}{abc}+ 1 \right )\geqq t+ \sqrt{3\,t} \,\,\,\,\left ( t= \frac{\left ( a+ b+ c \right )^{2}}{ab+ bc+ ca} \right )$$

(đẳng thức xảy ra khi $a= b= \left ( -4+ 3\sqrt{2} \right )c/\,2$)

 

 

 

$$\frac{1}{\left ( a+ b \right )^{2}}+ \frac{1}{\left ( b+ c \right )^{2}}+ \frac{1}{\left ( c+ a \right )^{2}}\geqq \frac{a+ b+ c}{4}$$

(Iran 96)

 

 

 

$$\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )\geqq 8$$

Bài viết đã được chỉnh sửa nội dung bởi DOTOANNANG: 11-07-2018 - 18:37

[DOTOANNANG]

$$\left ( a^{2}+ ab+ b^{2} \right )\left ( b^{2}+ bc+ c^{2} \right )\left ( c^{2}+ ca+ a^{2} \right )\geqq 27$$

 

 

 

$$\sqrt{2}\left ( a+ b+ c \right )\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )\geqq 8\left ( \sqrt{a^{2}+ bc}+ \sqrt{b^{2}+ ca}+ \sqrt{c^{2}+ ab} \right )$$

[DOTOANNANG]

$$ab\left ( a+ b \right )+ bc\left ( b+ c \right )+ ca\left ( c+ a \right )\geqq 2\left ( a+ b+ c \right )+ 2\sqrt{3}\left ( 1- abc \right )$$

[viet9a14124869]

 

$$\sqrt{2}\left ( a+ b+ c \right )\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )\geqq 8\left ( \sqrt{a^{2}+ bc}+ \sqrt{b^{2}+ ca}+ \sqrt{c^{2}+ ab} \right )$$

Gần tương tự ( Phạm Kim Hùng ) :

$$3(a+b+c)\geq 2(\sqrt{a^2+bc}+\sqrt{b^2+ca}+\sqrt{c^2+ab})$$

Bài viết đã được chỉnh sửa nội dung bởi viet9a14124869: 11-07-2018 - 18:11

[DOTOANNANG]

$$\frac{\left ( 2\,a+ b+ c \right )^{2}}{a\left ( 4\,a+ b+ c \right )}+ \frac{\left ( 2\,b+ c+ a \right )^{2}}{b\left ( 4\,b+ c+ a \right )}+ \frac{\left ( 2\,c+ a+ b \right )^{2}}{c\left ( 4\,c+ a+ b \right )}\geqq \frac{8}{abc}$$