Question

Доказать что в(в-4)>=-4​

Answer 1

$b(b-4)\geq -4\\\\b^2-4b+4\geq 0\\\\(b-2)^2\geq 0$

Answer 2

To prove that $v(v-4) \geq -4$ for all real values of $v$, we'll use algebraic manipulation. Here's the step-by-step proof:

Step 1: Expand the expression on the left-hand side (LHS).

$v(v-4) = v^2 - 4v$

Step 2: Rearrange the inequality.

$v^2 - 4v \geq -4$

Step 3: Move all terms to the left-hand side to get a quadratic inequality.

$v^2 - 4v + 4 \geq 0$

Step 4: Factor the quadratic expression.

$(v - 2)^2 \geq 0$

Step 5: Now, let's analyze the expression $(v - 2)^2$. The square of any real number is always greater than or equal to zero since both positive and negative numbers squared result in positive values, and zero squared is zero.

So, $(v - 2)^2 \geq 0$ for all real values of $v$.

Step 6: Since $(v - 2)^2 \geq 0$, we can conclude that $v^2 - 4v + 4 \geq 0$ for all real values of $v$.

Step 7: Finally, this inequality is equivalent to the original inequality $v(v-4) \geq -4$ for all real values of $v$.

Therefore, we have successfully proved that $v(v-4) \geq -4$ for all real values of $v$.