5 replies to this topic

[LoveMath1234]

Cho $a,b,c> 0$ sao cho: $a^4+b^4+c^4=3$

CMR: $a) \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3$

$b) \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq \frac{3}{2}$

[Master Kaiser]

Cho $a,b,c> 0$ sao cho: $a^4+b^4+c^4=3$

CMR: $a) \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3$

Ta có : $3(\sum a^2b)\leq (\sum a^2)^2=(\sum (\frac{a}{\sqrt{b}}.a\sqrt{b})^2)^2\leq VT.(\sum a^2b)$

$\Rightarrow VT\geq 3$

Edited by Master Kaiser, 10-02-2017 - 10:56.

[viet9a14124869]

Ta có : $3(\sum a^2b)\leq (\sum a^2)^2=(\sum (\frac{a}{\sqrt{b}}.a\sqrt{b})^2)^2\leq VT.(\sum a^2b)$

$\Rightarrow VT\geq 3$

tại sao lại viết dc thế này 

[Master Kaiser]

Cho $a,b,c> 0$ sao cho: $a^4+b^4+c^4=3$

CMR: $a) \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3$

$b) \frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq \frac{3}{2}$

 

Xin làm lại ...

 

Từ $a^4+b^4+c^4=3\Rightarrow a^2+b^2+c^2\geq a+b+c\Rightarrow \frac{a^2+b^2+c^2}{a+b+c}\geq 1$

 

a) Có : $\sum \frac{(a-b)^2}{b}\geq \frac{\sum (a-b)^2}{a+b+c}$

$\Leftrightarrow VT-(a+b+c)\geq \frac{3(a^2+b^2+c^2)}{a+b+c}-(a+b+c)$

$\Leftrightarrow VT\geq \frac{3(a^2+b^2+c^2)}{a+b+c}\geq 3$

 

b) Dễ dàng CM : $\sum \frac{a^3}{b+c}\geq \frac{a^2+b^2+c^2}{2}\Leftrightarrow \sum \left ( \frac{a^3}{b+c}+a^2 \right )\geq \frac{3(a^2+b^2+c^2)}{2}\Leftrightarrow VT(a+b+c)\geq \frac{3(a^2+b^2+c^2)}{2}$

$\Leftrightarrow VT\geq \frac{3(a^2+b^2+c^2)}{2(a+b+c)}\geq \frac{3}{2}$

 

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Dấu bằng khi a=b=c=1

[LoveMath1234]

Xin làm lại ...

 

Từ $a^4+b^4+c^4=3\Rightarrow a^2+b^2+c^2\geq a+b+c\Rightarrow \frac{a^2+b^2+c^2}{a+b+c}\geq 1$

 

a) Có : $\sum \frac{(a-b)^2}{b}\geq \frac{\sum (a-b)^2}{a+b+c}$

$\Leftrightarrow VT-(a+b+c)\geq \frac{3(a^2+b^2+c^2)}{a+b+c}-(a+b+c)$

$\Leftrightarrow VT\geq \frac{3(a^2+b^2+c^2)}{a+b+c}\geq 3$

 

b) Dễ dàng CM : $\sum \frac{a^3}{b+c}\geq \frac{a^2+b^2+c^2}{2}\Leftrightarrow \sum \left ( \frac{a^3}{b+c}+a^2 \right )\geq \frac{3(a^2+b^2+c^2)}{2}\Leftrightarrow VT(a+b+c)\geq \frac{3(a^2+b^2+c^2)}{2}$

$\Leftrightarrow VT\geq \frac{3(a^2+b^2+c^2)}{2(a+b+c)}\geq \frac{3}{2}$

 

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Dấu bằng khi a=b=c=1

Tại sao có cái này vậy?

[viet9a14124869]

Xin làm lại ...

 

Từ $a^4+b^4+c^4=3\Rightarrow a^2+b^2+c^2\geq a+b+c\Rightarrow \frac{a^2+b^2+c^2}{a+b+c}\geq 1$

 

a) Có : $\sum \frac{(a-b)^2}{b}\geq \frac{\sum (a-b)^2}{a+b+c}$

$\Leftrightarrow VT-(a+b+c)\geq \frac{3(a^2+b^2+c^2)}{a+b+c}-(a+b+c)$

$\Leftrightarrow VT\geq \frac{3(a^2+b^2+c^2)}{a+b+c}\geq 3$

 

b) Dễ dàng CM : $\sum \frac{a^3}{b+c}\geq \frac{a^2+b^2+c^2}{2}\Leftrightarrow \sum \left ( \frac{a^3}{b+c}+a^2 \right )\geq \frac{3(a^2+b^2+c^2)}{2}\Leftrightarrow VT(a+b+c)\geq \frac{3(a^2+b^2+c^2)}{2}$

$\Leftrightarrow VT\geq \frac{3(a^2+b^2+c^2)}{2(a+b+c)}\geq \frac{3}{2}$

 

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Dấu bằng khi a=b=c=1

Chỗ này sẽ sai vs a=b=$\frac{1}{2}$