Chủ đề này có 2 trả lời

[Haton Val]

​​Cho a,b,c dương . CMR : $\left ( 1+\frac{a}{b} \right )\left ( 1+\frac{b}{c} \right )\left ( 1+\frac{c}{a} \right )\geq 2\left ( 1+\frac{a+b+c}{\sqrt[3]{abc}} \right )$

[cyndaquil]

Vì $\left ( 1+\frac ab \right )\left ( 1+\frac bc \right )\left ( 1+\frac ca \right )=(a+b+c)\left( \frac 1a+\frac 1b+\frac 1c\right)-1$

Nên chỉ cần chứng minh 

$(a+b+c)\left( \frac 1a+\frac 1b+\frac 1c\right) \ge 3+\frac{2(a+b+c)}{\sqrt[3]{abc}}$

Ta có 

$\frac 13(a+b+c)\left( \frac 1a+\frac 1b+\frac 1c\right)  \ge 3$

$\frac 23(a+b+c)\left( \frac 1a+\frac 1b+\frac 1c\right)  \ge \frac 23(a+b+c).\frac{3}{\sqrt[3]{abc}}=\frac{2(a+b+c)}{\sqrt[3]{abc}}$

Cộng 2 vế suy ra dpcm

[tienduc]

​​Cho a,b,c dương . CMR : $\left ( 1+\frac{a}{b} \right )\left ( 1+\frac{b}{c} \right )\left ( 1+\frac{c}{a} \right )\geq 2\left ( 1+\frac{a+b+c}{\sqrt[3]{abc}} \right )$

Cách khác

$VT=(1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})=2+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{c}{b}=(\frac{a}{b}+\frac{a}{c}+\frac{a}{a})+(\frac{b}{c}+\frac{b}{a}+\frac{b}{b})+(\frac{c}{a}+\frac{c}{b}+\frac{c}{c})-1$

Áp dụng BĐT $Cauchy$ cho 3 số ta có

$\frac{a}{b}+\frac{a}{c}+\frac{a}{a}\geq \frac{3a}{\sqrt[3]{abc}};\frac{b}{c}+\frac{b}{a}+\frac{b}{b}\geq \frac{3b}{\sqrt[3]{abc}};\frac{c}{a}+\frac{c}{b}+\frac{c}{c}\geq \frac{3c}{\sqrt[3]{abc}}$

Cộng vế $\rightarrow VT\geq 2(\frac{a+b+c}{\sqrt[3]{abc}})+\frac{a+b+c}{\sqrt[3]{abc}}-1\geq 2(\frac{a+b+c}{\sqrt[3]{abc}})+\frac{3\sqrt[3]{abc}}{\sqrt[3]{abc}}-1= 2(\frac{a+b+c}{\sqrt[3]{abc}})+2=VP$

Bài viết đã được chỉnh sửa nội dung bởi tienduc: 23-03-2017 - 12:25