Question:

Step 1: Determine the value of $a$. The expression for $y$ is given by $y = \dfrac{(x-3)^2 + 3}{x - 2}$. Let $t = x - 2$. Then $x - 3 = t - 1$. Substituting these into the expression for $y$: $$y = \frac{(t-1)^2 + 3}{t} = \frac{t^2 - 2t + 1 + 3}{t} = \frac{t^2 - 2t + 4}{t} = t - 2 + \frac{4}{t}$$ Step 2: Find the least positive value of $y$. For $y$ to be least positive, $t$ must be positive, which implies $x - 2 > 0$, or $x > 2$. By the AM-GM inequality, for $t > 0$: $$t + \frac{4}{t} \geq 2\sqrt{t \cdot \frac{4}{t}} = 2\sqrt{4} = 4$$ Equality holds when $t = \frac{4}{t}$, which implies $t^2 = 4$. Since $t > 0$, we must have $t = 2$. Substituting $t=2$ back into the expression for $y$: $$y = 2 - 2 + \frac{4}{2} = 0 + 2 = 2$$ The least positive value of $y$ is $2$, which occurs when $t = 2$. Since $t = x - 2$, we have $2 = x - 2$, which yields $x = 4$. The problem states that the value of $x$ is equal to $a$, so $a = 4$. Step 3: Determine the relationship between $a, b, c$. The condition $a^3 + b^3 + c^3 \leq 3abc$ is given for $a, b, c > 0$. By the AM-GM inequality, for $a, b, c > 0$: $$a^3 + b^3 + c^3 \geq 3\sqrt[3]{a^3 b^3 c^3} = 3abc$$ For the given condition $a^3 + b^3 + c^3 \leq 3abc$ to hold true simultaneously with the AM-GM inequality $a^3 + b^3 + c^3 \geq 3abc$, it must be that: $$a^3 + b^3 + c^3 = 3abc$$ This equality holds if and only if $a = b = c$. Since $a = 4$, it follows that $b = 4$ and $c = 4$. Step 4: Evaluate the given statements using $a=b=c=4$. **Statement 1:** $\log_2(a+b+c) = \log_2(abc)$ Substituting $a=b=c=4$: LHS: $\log_2(4+4+4) = \log_2(12)$ RHS: $\log_2(4 \cdot 4 \cdot 4) = \log_2(64)$ Since $\log_2(12) \neq \log_2(64)$, this statement is false. Alternatively, the statement implies $a+b+c = abc$. With $a=b=c$, this simplifies to $3a = a^3$, or $a^2 = 3$. This means $a = \sqrt{3}$. Since $a=4 \neq \sqrt{3}$, the statement is false. **Statement 2:** $a+b < c$ Substituting $a=b=c=4$: $4+4 < 4 \implies 8 < 4$ This statement is false. **Statement 3:** $\log_2 a + \log_2 b + \log_2 c = 3\log_2 a$ Substituting $a=b=c=4$: LHS: $\log_2 4 + \log_2 4 + \log_2 4 = 2 + 2 + 2 = 6$ RHS: $3\log_2 4 = 3 \cdot 2 = 6$ Since LHS = RHS, this statement is true. Alternatively, using logarithm properties, $\log_2 a + \log_2 b + \log_2 c = \log_2(abc)$. Since $a=b=c$, this becomes $\log_2(a^3) = 3\log_2 a$, which is an identity for $a>0$. Thus, the statement is true. **Statement 4:** $\dfrac{b^2 + 3a}{4} = 7$ Substituting $a=b=c=4$: LHS: $\dfrac{4^2 + 3(4)}{4} = \dfrac{16 + 12}{4} = \dfrac{28}{4} = 7$ RHS: $7$ Since LHS = RHS, this statement is true. Therefore, Statement 3 and Statement 4 are true.