Which of the following is correct? A: for a > 0, a³ + 1/a² has minimum 2√a. B: for x > 1, x + 1/x has minimum 2. C: for x > 0, √x + 1/√x ≥ 2. D: for x < 5/4, y = 4x − 2 + 1/(4x − 5) has minimum 1.

Step-by-step solution

1. 1

   Recall the three conditions AM–GM needs. For $u+v\ge 2\sqrt{uv}$ to give a genuine minimum, the terms must be positive, their product must be constant, and the equality point $u=v$ must actually lie in the allowed domain. Each wrong option below fails exactly one of these.

2. 2

   A fails: the bound is not a constant. AM–GM gives $a^{3}+\dfrac{1}{a^{2}}\ge 2\sqrt{a^{3}\cdot a^{-2}}=2\sqrt{a}$, but $2\sqrt a$ still depends on $a$, so it is not a minimum value — it is a moving target. (The true minimum is at $3a^{2}=2a^{-3}$, i.e. $a=(2/3)^{1/5}$.) Incorrect.

3. 3

   B fails: equality is excluded by the domain. $x+\dfrac1x\ge 2$ with equality only at $x=1$, and the domain is $x>1$. On $x>1$ the function is strictly increasing, so it takes values $>2$ but never reaches $2$: the infimum is $2$ and there is no minimum. Incorrect.

4. 4

   C is correct. Put $t=\sqrt{x}>0$; then $t+\dfrac1t\ge 2\sqrt{t\cdot\tfrac1t}=2$, with equality at $t=1$, i.e. $x=1$, which is in the domain $x>0$. So the inequality holds for all $x>0$ and is sharp. Correct.

5. 5

   D fails: 1 is a maximum, not a minimum. Write $t=4x-5$; since $x<\tfrac54$, $t<0$. Then $4x-2=t+3$, so

   $y=t+3+\frac1t=3+\left(t+\frac1t\right)$

   For $t<0$, $t+\dfrac1t=-\left(|t|+\dfrac{1}{|t|}\right)\le -2$, hence $y\le 1$. So $1$ is the largest value of $y$ (attained at $t=-1$, i.e. $x=1$), not the smallest. Incorrect.

6. 6

   Conclude.

   $\boxed{\text{C}}$

   Spot checks: for D, $x=1$ gives $y=4-2+\tfrac{1}{-1}=1$, and $x=0$ gives $y=-2-\tfrac15=-2.2<1$, confirming that $1$ is a maximum. For B, $x=1.1$ gives $1.1+0.909=2.009>2$, and the value decreases toward $2$ but never reaches it.