Understanding the intuition behind a proof of Holder’s Inequality I’m reading a book on Linear Algebra and one of the exercises is to prove Holder’s inequality. In the exercise, the author actually breaks the question in several subitens I order to help the reader. The very first item to prove is the following:
a) Consider the function $f(t) = (1-\lambda) + \lambda t - t^\lambda$, where $\lambda \in (0,1)$. Use it to prove that:
$$
\alpha^\lambda \beta ^{1-\lambda} \leq \lambda \alpha + (1-\lambda)\beta.
$$
The inequality can be proved by studying the sign of $f(t)$ and using $t=\alpha/\beta$. My question though is “where does the function $f(t)$ comes from?”. I mean, I’m proving Holder’s inequality, but it’s not clear the intuition behind this proof, such as from where this $f(t)$ was intuited from.
The rest of the exercise to prove Holder comes from using (a) with $\frac{x}{||x||}$ and $\frac{y}{||y||}$.
 A: The intuition behind $f(t)$ is quite simple.
Note that we are dealing with a homogeneous function. A function $g$ is called homogeneous of degree $p\in\mathbb{R}$ if $g(cx, cy) = c^p g(x,y) \ \forall c,x,y$. For example $x^2 + 2y^2 - xy$ and $\frac{1}{x} - \frac{2}{y} - \frac{3}{\sqrt{xy}}$ are homogeneous of degree $2$ and $-1$, respectively.
Suppose we want to prove
$$g(x,y)\ge 0 \quad \forall x > 0, y > 0,$$
where $g$ is homogeneous of degree $p$ for some $p$. It suffices to set one of variables to $1$. Why? Notice that
$$g(x,y) = y^p g(x/y,1) = x^p g(1, y/x),$$
the original inequality is equivalent to both $g(x/y,1) > 0$ and $g(1, y/x) > 0$ for any $x,y$, i.e., it suffices to prove that $g(t,1) > 0$ or alternatively $g(1, t) > 0$ for any $t$, which is the original inequality when we set one of the variables to $1$.
The above explanation looks complicated but the rule is very simple: whenever you have a homogeneous inequality (with positive variables), just divide both sides by either $y^p$ and use $t=x/y$ as the new variable (or alternatively divide by $x^p$ and use $t=y/x$).
Let's take your above example:
$$\lambda \alpha + (1-\lambda)\beta - \alpha^\lambda \beta ^{1-\lambda} \ge 0.$$
Divide by $\beta > 0$ the above becomes $\lambda t + (1-\lambda) - t^{\lambda} \ge 0$ where $t=\alpha/\beta$, which is where your $f$ comes from. Alternatively, divide by $\alpha > 0$ it becomes $\lambda + (1-\lambda)t - t^{1-\lambda}\ge 0$ where $t=\beta/\alpha$, which means you can consider a different function $f$.

Bonus: How to deal with real-valued variables instead of positive ones? Just reduce it to the positive case. For example, the inequality $x^2 + y^2\ge 2xy$ holds for real $x,y$. It suffices to prove it for non-negative $x,y$ because then $x^2 + y^2\ge 2|x||y| \ge 2xy$. Next, if one of the variables is $0$ then the inequality is clearly true. It remains to prove it for positive $x,y$.
