Using a sequence of natural numbers and AM-GM inequality, I have shown
$$\eta x_1 + (1 - \eta)x_2 \geq x_1^{\eta}x_2^{1-\eta}$$
holds for $\eta \in [0,1]$, $x_1,x_2 \in \mathbb{R}$. Now I need to prove that the equality holds if and only if $x_1 = x_2$ but am having a trouble in doing so(Of course, the if part is trivial).
I have tried having a fixed $\eta$ and $x_1$, and expressing $x_2 = x_1 + \epsilon$ to turn the above equation into a univariate function, but could not figure out a way to show the desired property.
How do I show the above equation without using the convexity of $-\mathrm{log}(x)$? I need to avoid using logarithm because I am proving the above condition to prove the strong convexity of logarithms.
You could continue on your approach using Bernoulli's Inequality. For ease, I will change notation slightly, WLOG, let $\epsilon \geqslant 0$ and $$x_2 = x_1(1+\epsilon) \implies x_2^{1-\eta} = x_1^{1-\eta}(1+\epsilon)^{1-\eta} \leqslant x_1^{1-\eta}(1+\epsilon-\epsilon\eta)$$ $$\implies x_1^\eta x_2^{1-\eta} \leqslant x_1^\eta \cdot x_1^{1-\eta} (1+\epsilon-\epsilon \eta) = x_1\left(\eta +(1-\eta)(1+\epsilon)\right) = \eta x_1 + (1-\eta)x_2$$
Equality for the Bernoulli's inequality used is possible iff $\eta \in \{0, 1\}$ or $\epsilon=0$, thereby giving the condition $x_1=x_2$.