1)
BPT tương đương:
$a+b\geq2\sqrt{ab}$
$⇔a+b-2\sqrt{ab}\geq0$
$⇔\left(\sqrt{a}-\sqrt{b}\right)²\geq0$
Dấu đẳng thức xảy ra khi $a=b$
2)
+) $(a-b)^2 ≥ 0$
$⇔ a^2 + 2ab + b^2 ≥ 4ab$
$⇔ (a+b)^2 ≥ 4ab$
$⇔ a+b ≥ 2\sqrt{ab}$
Áp dụng BĐT ta có:
$⇒ a+\dfrac{1}{a} \ge 2\sqrt{a.\dfrac{1}{a}}$
$⇔ a+\dfrac{1}{a} \ge 2$ (đpcm)
Dấu đẳng thức xảy ra khi $a=1$
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Cho $a$ và $b$ dương, chứng minh:
$\left(a+b\right)\left(1+ab\right)\ge4ab$
$⇔ \dfrac{\left(a+b\right)\left(1+ab\right)}{ab}\ge4$
$⇔ \left(ab+1\right).\dfrac{a+b}{ab}\ge4$
$⇔ \left(ab+1\right).\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4$
$⇔\begin{cases}ab+1\ge2\sqrt{ab}\\\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}\end{cases}$
$⇔\left(ab+1\right).\left(\dfrac{1}{a}+\dfrac{1}{b}\right)≥2\sqrt{ab}.2\sqrt{\dfrac{1}{ab}}$
$⇔\left(ab+1\right).\left(\dfrac{1}{a}+\dfrac{1}{b}\right)≥4\sqrt{ab.\dfrac{1}{ab}}$
$⇔\left(ab+1\right).\left(\dfrac{1}{a}+\dfrac{1}{b}\right)≥4$ (đpcm)
