Cho $a,b,c>0$. CMR:
$a) \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b+\frac{4(a-b)^2}{a+b}$
$b) \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-c)^2}{a+b+c}$
Cho $a,b,c>0$. CMR:
$a) \frac{a^{2}}{b}+\frac{b^{2}}{a}\geq a+b+\frac{4(a-b)^2}{a+b}$
$b) \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^2}{a}\geq a+b+c+\frac{4(a-c)^2}{a+b+c}$
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:
$$(\frac{a^2}{b}+b-2a)+(\frac{b^2}{a}+a-2b) \geq \frac{4(a-b)^2}{a+b}$$
$$\frac{(a-b)^2}{b}+\frac{(a-b)^2}{a} \geq \frac{4(a-b)^2}{a+b}$$
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
\[\text{VT - VP} = \frac{a(b^2-ca)^2+c(a^2-2ab+bc)^2+b(ab-2ca+c^2)^2}{abc(a+b+c)} \geqslant 0.\]
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:
$\left ( a+ b+ c \right )\sum \frac{\left ( a- b \right )^{2}}{b}\geq 4. max \left ( \left | a- b \right |, \left | b- c \right |, \left | c- a \right | \right )^{2}$
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