Chủ đề này có 3 trả lời

[melodias2002]

Cho $a,b,c>0$. CMR:

$a) \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b+\frac{4(a-b)^2}{a+b}$

$b) \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-c)^2}{a+b+c}$ 

[nmtuan2001]

Cho $a,b,c>0$. CMR:

$a) \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b+\frac{4(a-b)^2}{a+b}$

$b) \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-c)^2}{a+b+c}$ 

$$\frac{a^2}{b}+b-2a=\frac{a^2+b^2-2ab}{b}=\frac{(a-b)^2}{b}$$

 

$a)$ BĐT tương đương với:

$$(\frac{a^2}{b}+b-2a)+(\frac{b^2}{a}+a-2b) \geq \frac{4(a-b)^2}{a+b}$$

$$\frac{(a-b)^2}{b}+\frac{(a-b)^2}{a} \geq \frac{4(a-b)^2}{a+b}$$

$$\frac{1}{a}+\frac{1}{b} \geq \frac{4}{a+b}$$

BĐT đúng theo AM-GM: $(a+b)(\frac{1}{a}+\frac{1}{b}) \geq 2\sqrt{ab}.2\sqrt{\frac{1}{ab}}=4$.

 

$b)$ BĐT tương đương với:
$$(\frac{a^2}{b}+b-2a)+(\frac{b^2}{c}+c-2b)+(\frac{c^2}{a}+a-2c) \geq \frac{4(a-c)^2}{a+b+c}$$

$$\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a} \geq \frac{4(a-c)^2}{a+b+c}$$

BĐT đúng theo Cauchy-Schwarz: $VT=\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(a-c)^2}{a} \geq \frac{(a-b+b-c+a-c)^2}{b+c+a}=\frac{4(a-c)^2}{a+b+c}=VP$.

[Nguyenhuyen_AG]

Cho $a,b,c>0$. CMR:

$b) \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-c)^2}{a+b+c}$ 

 

Ta có

\[\text{VT - VP} = \frac{a(b^2-ca)^2+c(a^2-2ab+bc)^2+b(ab-2ca+c^2)^2}{abc(a+b+c)} \geqslant 0.\]

[DOTOANNANG]

Viết lại BĐT:
$\sum \frac{\left ( a- b \right )^{2}}{b}\geq \frac{4\left ( a- b \right )^{2}}{a+ b+ c}$

Dẫn tới:

$\left ( a+ b+ c \right )\sum \frac{\left ( a- b \right )^{2}}{b}\geq 4\left ( a- b \right )^{2}$

Điều này đúng vì:

$\left ( a+ b+ c \right )\sum \frac{\left ( a- b \right )^{2}}{b}\geq 4. max \left ( \left | a- b \right |, \left | b- c \right |, \left | c- a \right | \right )^{2}$