[Toán 10] Bất đẳng thức

N

noinhobinhyen

sử dụng AM-GM ngược dấu

Ta có $\dfrac{a^3}{3a^2+b} = \dfrac{a}{3}-\dfrac{ab}{3(3a^2+b)}$

$\geq \dfrac{1}{3}.[a-\dfrac{ab}{2\sqrt{3a^2b}}] = \dfrac{1}{3}[a-\dfrac{\sqrt{b}}{2\sqrt{3}}]$

$\Rightarrow P \geq \dfrac{1}{3}(a+b+c)-\dfrac{1}{6\sqrt{3}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

Do $(\sqrt{a}+\sqrt{b}+\sqrt{c}) \leq \sqrt{3(a+b+c)}$

$\Rightarrow P \geq \dfrac{1}{3}(a+b+c)-\dfrac{\sqrt{a+b+c}}{6}$

Vậy ta đặt $t=\sqrt{a+b+c} \geq 1$

$\Rightarrow P \geq \dfrac{2t^2-t}{6}$

$\Rightarrow MIN P = \dfrac{1}{6} \Leftrightarrow t=1$
 
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