Question

Доказать: a - 3√ab + 4b ≥ √ab​

Answer 1

Пусть: a>=0 ; b>=0

a-3√ab +4b>=√ab

a-4√ab +4b>=0

тк  a>=0 ; b>=0 , то  существуют квадратные корни: √a и √b

Тогда это неравенство можно представить  в виде:

(√a)^2 - 2*√a *(2√b) +(2*√b)^2>=0

(√a-2√b)^2>=0 ,тк  квадрат  числа неотрицателен.

ЧТД.

Answer 2

To prove the inequality $a - 3\sqrt[3]{ab} + 4b \geq \sqrt{ab}$, we need to use the properties of inequalities and basic algebraic manipulations. Let's proceed step by step:

Step 1: Simplify the expression on the left side of the inequality. $a - 3\sqrt[3]{ab} + 4b$

Step 2: Get a common denominator for the terms involving $\sqrt[3]{ab}$ and $b$ so that we can compare the two sides of the inequality effectively. $\sqrt{ab} = \sqrt[3]{ab} \cdot \sqrt[3]{ab}$

Step 3: Now, rewrite the left-hand side of the inequality using the common denominator $\sqrt{ab}$. $a - 3\sqrt[3]{ab} + 4b = \sqrt[3]{ab} \cdot \sqrt[3]{ab} - 3\sqrt[3]{ab} + 4b$

Step 4: Combine like terms on the left-hand side of the inequality. $\sqrt[3]{ab} \cdot \sqrt[3]{ab} - 3\sqrt[3]{ab} + 4b = \sqrt[3]{ab}(\sqrt[3]{ab} - 3) + 4b$

Now we have the inequality in the following form: $\sqrt[3]{ab}(\sqrt[3]{ab} - 3) + 4b \geq \sqrt{ab}$

Step 5: Now, let's assume that $a$ and $b$ are positive real numbers. In that case, $\sqrt[3]{ab} > 0$ and $\sqrt{ab} > 0$. Thus, we can divide both sides of the inequality by $\sqrt[3]{ab}$ without changing the inequality sign.

$\frac{\sqrt[3]{ab}(\sqrt[3]{ab} - 3) + 4b}{\sqrt[3]{ab}} \geq \frac{\sqrt{ab}}{\sqrt[3]{ab}}$

Step 6: Simplify both sides.

$\sqrt[3]{ab} - 3 + \frac{4b}{\sqrt[3]{ab}} \geq \sqrt[3]{ab^2}$