Let $a,b\in\mathbb{R},~a>1,~b>0.$ Find the least possible value for $\alpha$ such that :$$(a+b)^x\geq a^x+b,~(\forall)x\geq\alpha.$$
1. We need to find the least possible value for $\alpha$ such that the inequality 
   \[
   (a+b)^x \geq a^x + b
   \]
   holds for all $x \geq \alpha$, given that $a > 1$ and $b > 0$.

2. Let's start by testing $\alpha = 1$. We need to check if the inequality holds for $x = 1$:
   \[
   (a+b)^1 \geq a^1 + b
   \]
   Simplifying, we get:
   \[
   a + b \geq a + b
   \]
   This is true, so $\alpha = 1$ is a candidate.

3. Next, we need to show that the function $f(x) = (a+b)^x - a^x - b$ is increasing for $x \geq 1$. We will do this by showing that the derivative $f'(x)$ is non-negative for $x \geq 1$.

4. Compute the derivative of $f(x)$:
   \[
   f'(x) = \frac{d}{dx} \left( (a+b)^x - a^x - b \right)
   \]
   Using the chain rule, we get:
   \[
   f'(x) = (a+b)^x \ln(a+b) - a^x \ln(a)
   \]

5. We need to show that $f'(x) \geq 0$ for $x \geq 1$. This means:
   \[
   (a+b)^x \ln(a+b) \geq a^x \ln(a)
   \]

6. Since $a > 1$ and $b > 0$, we have $a+b > a > 1$. Therefore, $\ln(a+b) > \ln(a) > 0$. Also, $(a+b)^x > a^x$ for $x \geq 1$ because $a+b > a$.

7. Thus, $(a+b)^x \ln(a+b) > a^x \ln(a)$ for $x \geq 1$, which implies $f'(x) \geq 0$ for $x \geq 1$. Therefore, $f(x)$ is increasing for $x \geq 1$.

8. Since $f(x)$ is increasing for $x \geq 1$ and $f(1) = 0$, it follows that $f(x) \geq 0$ for $x \geq 1$. This means:
   \[
   (a+b)^x \geq a^x + b \quad \text{for all} \quad x \geq 1
   \]

9. To confirm that $\alpha = 1$ is the least possible value, assume $\alpha < 1$. Let $x = 1 - \epsilon$ for some small $\epsilon > 0$. We need to check if the inequality holds:
   \[
   (a+b)^{1-\epsilon} \geq a^{1-\epsilon} + b
   \]

10. Using the binomial approximation for small $\epsilon$, we have:
    \[
    (a+b)^{1-\epsilon} \approx (a+b) \left(1 - \epsilon \ln(a+b)\right)
    \]
    and
    \[
    a^{1-\epsilon} \approx a \left(1 - \epsilon \ln(a)\right)
    \]

11. Substituting these approximations, we get:
    \[
    (a+b) \left(1 - \epsilon \ln(a+b)\right) \geq a \left(1 - \epsilon \ln(a)\right) + b
    \]

12. Simplifying, we get:
    \[
    a + b - \epsilon (a+b) \ln(a+b) \geq a - \epsilon a \ln(a) + b
    \]

13. Canceling $a + b$ from both sides, we get:
    \[
    -\epsilon (a+b) \ln(a+b) \geq -\epsilon a \ln(a)
    \]

14. Dividing by $-\epsilon$ (which is positive), we get:
    \[
    (a+b) \ln(a+b) \leq a \ln(a)
    \]

15. Since $a > 1$ and $b > 0$, we have $a+b > a > 1$ and $\ln(a+b) > \ln(a) > 0$. Therefore, $(a+b) \ln(a+b) > a \ln(a)$, which contradicts the inequality. Hence, $\alpha < 1$ cannot be the least possible value.

16. Therefore, the least possible value for $\alpha$ is $\alpha = 1$.

The final answer is $\boxed{1}$