Proof inequality using convexity I struggling with proofing an inequality. We have to show that
$x - y \le (1-\theta)^{-1} x^\theta (x^{1-\theta} - y^{1-\theta})$
holds for all $x, y > 0, \theta \in [0, 1)$. Further we know that $f(x) = x^\frac{1}{1-\theta}$ is convex on $[0, \infty)$.
I tried to use the first order condition of convexity (see here). With $f(x)$ as mentioned above and $f'(x) = (1 - \theta)^{-1} x^\frac{\theta}{1-\theta} $ we get
$y^\frac{1}{1-\theta} \ge x^\frac{1}{1-\theta} + (1 - \theta)^{-1} x^\frac{\theta}{1-\theta} (y-x)$
If I try to rearrange the terms, the closest I get is
$x^\frac{-\theta}{1-\theta} (y^\frac{1}{1-\theta} - x^\frac{1}{1-\theta}) \ge (1 - \theta)^{-1}  (y-x)$
$(x^\frac{-\theta}{1-\theta} y^\frac{1}{1-\theta} - x^\frac{1 - \theta}{1-\theta}) \ge (1 - \theta)^{-1}  (y-x)$
$(x^\frac{-\theta}{1-\theta} y^\frac{1}{1-\theta} - x) \ge (1 - \theta)^{-1}  (y-x)$
$x^\theta (x^\frac{-2\theta + \theta^2}{1-\theta} y^\frac{1}{1-\theta} - x^{1-\theta}) \ge (1 - \theta)^{-1}  (y-x)$
Now I don't know what I should do next, and anyway, is it the right way to begin?
 A: Consider the function $$F(x,y,\theta) := x^\theta \frac{x^{1-\theta} - y^{1-\theta}}{1-\theta} - (x-y).$$
The claim is equivalent to showing that $F(x,y,\theta) \geq 0$ for all $x,y,\theta$. It is easy to spot that $F(x,x,\theta) = 0$, so we hope this should be the mininum. What is more, the inequality does not change if we scale both $x$ and $y$. If put $y = e^t x$, then we have:
$$ F(x,y,\theta) = x \left( 1-e^t - \frac{1-e^{t(1-\theta)}}{(1-\theta)} \right) = x G(t,\theta).$$ 
So, it will suffice to show that $G(t,\theta)$ is always positive. The simplest solution seems to be to fix $\theta$, and constider the function $g(t) = G(t,\theta)$. It is easy to notice that:
$$ g'(t) = e^t - e^{t(1-\theta)} = e^{t(1-\theta)}(e^{t \theta} - 1). $$
Looking at the signs $g'(t)$ takes, we conclude that it has just one minimum at $t=0$. The value $g(0)$ is just $0$ (this is the same $0$ as the one in $F(x,x,\theta)$). So, $g(t) \geq 0$, and conseguently $G\geq 0$, and $F \geq 0$, and the claim follows.
