Toán 9 bất đẳng thức 9

[Thu Anh 14305]

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[tex]2\left ( \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \right )\geq 1+\frac{b}{b+2a}+\frac{c}{c+2a}+\frac{a}{a+2c}[/tex]
sử dụng bđt gợi ý là: [tex]\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}[/tex] nha các bạn
giúp mình với mình cảm ơn nhiều lắm nhé

 

[Kaito Kidㅤ]

[tex]\frac{1}{b+2c}+\frac{1}{b+2a}\geq \frac{4}{2a+2b+2c}=\frac{2}{a+b+c}\\\rightarrow \frac{a}{b+2c}+\frac{a}{b+2a}\geq \frac{2a}{a+b+c}(1)\\TT:\frac{b}{c+2a}+\frac{b}{c+2b}\geq \frac{2b}{a+b+c}(2)\\\frac{c}{a+2c}+\frac{c}{a+2b}\geq \frac{2c}{a+b+c}(3)\\(1)+(2)+(3):\\\rightarrow \frac{a}{b+2c}+\frac{a}{b+2a}+\frac{b}{c+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}+\frac{c}{a+2b}\geq 2\\\rightarrow \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 2-\frac{a}{b+2a}-\frac{b}{c+2b}-\frac{c}{a+2c}\\\rightarrow 2(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b})\geq 4-\frac{2a}{b+2a}-\frac{2b}{c+2b}-\frac{2c}{a+2c}=1+1-\frac{2a}{b+2a}+1-\frac{2b}{c+2b}+1-\frac{2c}{a+2c}=1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}(dpcm)[/tex]

 

[Thu Anh 14305]

[tex]\frac{1}{b+2c}+\frac{1}{b+2a}\geq \frac{4}{2a+2b+2c}=\frac{2}{a+b+c}\\\rightarrow \frac{a}{b+2c}+\frac{a}{b+2a}\geq \frac{2a}{a+b+c}(1)\\TT:\frac{b}{c+2a}+\frac{b}{c+2b}\geq \frac{2b}{a+b+c}(2)\\\frac{c}{a+2c}+\frac{c}{a+2b}\geq \frac{2c}{a+b+c}(3)\\(1)+(2)+(3):\\\rightarrow \frac{a}{b+2c}+\frac{a}{b+2a}+\frac{b}{c+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}+\frac{c}{a+2b}\geq 2\\\rightarrow \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 2-\frac{a}{b+2a}-\frac{b}{c+2b}-\frac{c}{a+2c}\\\rightarrow 2(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b})\geq 4-\frac{2a}{b+2a}-\frac{2b}{c+2b}-\frac{2c}{a+2c}=1+1-\frac{2a}{b+2a}+1-\frac{2b}{c+2b}+1-\frac{2c}{a+2c}=1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}(dpcm)[/tex]

cảm ơn bạn nhiều lắm nhé!