Question

The smallest integral value of $\alpha$ for which the inequality $1 + log_4(x^2 + 1) \leq log_4(\alpha x^2 + 2x + \alpha)$ holds true for all $x \in R$ is ______

Answer 1

5

Answer 2

We are given:

$$1 + \log_4(x^2+1) \le \log_4(\alpha x^2+2x+\alpha)$$

Since $\log_4$ is an increasing function, the inequality is equivalent to:

$$4(x^2+1) \le \alpha x^2 + 2x + \alpha \quad \forall\, x \in \mathbb{R}.$$

Rearranging, we have:

$$\alpha x^2 + 2x + \alpha - 4x^2 - 4 \ge 0,$$

which simplifies to:

$$(\alpha-4)x^2 + 2x + (\alpha-4) \ge 0.$$

Let $A = \alpha - 4$. Then the inequality becomes:

$$Ax^2 + 2x + A \ge 0 \quad \forall\, x.$$

For this quadratic to be non-negative for all $x$ (and considering $A>0$), its discriminant must be non-positive:

$$\Delta = 2^2 - 4A^2 \le 0 \quad \Longrightarrow \quad 4 - 4A^2 \le 0,$$ $$\Longrightarrow \quad 1 - A^2 \le 0 \quad \Longrightarrow \quad A^2 \ge 1.$$

Since $A = \alpha - 4 > 0$ (because the quadratic must be convex), we require:

$$\alpha - 4 \ge 1 \quad \Longrightarrow \quad \alpha \ge 5.$$

Thus, the smallest integral value of $\alpha$ is 5.