Proving an convexity-looking inequality 
If $0 \le \alpha \le 1$ and $0 \le \lambda \le 1$, then
  $$\lambda^\alpha x^\alpha +(1-\lambda^\alpha) y^\alpha \ge (\lambda x + (1-\lambda)y)^\alpha$$
  whenever $0 \le y \le x$.

This looks like a standard sort of inequality that resembles convexity... if anyone can provide some "big picture" context, I would really appreciate it.
I've tried to prove it below, but it seems rather cumbersome. If there is a faster way, please  let me know.

If $\alpha=0$ or $\alpha = 1$, then the lemma holds. If $\lambda=0$ or $\lambda=1$, then the lemma holds. If $x=0$, then $y=0$, so the lemma also holds. So, we assume $0 < \alpha < 1$, $0 < \lambda < 1$, and $x > 0$.
Note that for $0<\alpha \le 1$ and $0 < \lambda \le 1$ in the given range, we have $$1-\lambda^\alpha \ge (1-\lambda)^\alpha.$$ To see this, note that for fixed $\alpha$, the function $1-\lambda^\alpha - (1-\lambda)^\alpha$ has zeros at $\lambda=0$ and $\lambda=1$ and a single critical point at $\lambda=1/2$, which is a local minimum because $\alpha < 1$. Therefore, if we show
$$\lambda^\alpha x^\alpha + (1-\lambda)^\alpha y^\alpha \ge (\lambda x + (1-\lambda)y)^\alpha,$$
(replacing $1-\lambda^\alpha$ with $(1-\lambda)^\alpha$), we will have proven the lemma.
Let $f(y)=\lambda^\alpha x^\alpha + (1-\lambda)^\alpha y^\alpha - (\lambda x + (1-\lambda)y)^\alpha$. Then,
$$f'(y) = (1-\lambda)^\alpha \alpha y^{\alpha-1} -\alpha(1-\lambda)(\lambda x + (1-\lambda)y)^{\alpha-1}.$$
Setting $f'(y)=0$ implies
\begin{align*}
(1-\lambda)^\alpha y^{\alpha-1} &= (1-\lambda)(\lambda x + (1-\lambda)y)^{\alpha-1}\\
((1-\lambda)y)^{\alpha-1} &= (\lambda x + (1-\lambda)y)^{\alpha-1}\\
\lambda x &= 0.
\end{align*}
Since we know $\lambda x \ne 0$ by assumption, we have a contradiction, so there is no critical point. Therefore $f$ is either increasing or decreasing for $y \ge 0$. Since $f(0)=0$ and
\begin{align*}
f(x)&=\lambda^\alpha x^\alpha + (1-\lambda)^\alpha x^\alpha - x^\alpha\\
&=x^\alpha (\lambda^\alpha + (1-\lambda)^\alpha - 1)\\
&\ge x^\alpha (\lambda + (1-\lambda)-1) & \text{$0<\alpha<1$ and $0<\lambda<1$ imply $\lambda^\alpha \ge \lambda$}\\
&=0,
\end{align*}
we see that $f(y)$ is increasing for $0\le y \le x$, and therefore nonnegative.
 A: We will prove the equivalent inequality $$(\lambda x + (1-\lambda)y)^\alpha  - (\lambda x)^\alpha \leq (\lambda y + (1-\lambda)y)^\alpha -(\lambda y)^\alpha$$ which is obvious because $x^\alpha$ is concave and $\lambda x \geq \lambda y$, then when we add $(1-\lambda)y$ to  both $\lambda x$ and $\lambda y$ we know which increment is bigger 
A: Note that when $0<p<1$ the function $x\mapsto x^{p}$ is subadditive on the non-negative reals so:
$$(\lambda x+(1-\lambda)y)^{\alpha}\le(\lambda x)^{\alpha}+((1-\lambda)y)^{\alpha}=\lambda^{\alpha}x^{\alpha}+(1-\lambda)^{\alpha}y^{\alpha}$$
To show that $x^{p}$ is subadditive first note that
$$(x+y)^{p}\le x^{p}+y^{p}$$
is clear when $y=0$ and hence if we can show it is true when $y=1$ then:
$$(x+y)^{p}=y^{p}(\frac{x}{y}+1)^{p}\le y^{p}((\frac{x}{y})^{p}+1)=x^{p}+y^{p}$$
Now just show that $f(x)=(x+1)^{p}-x^{p}$ is decreasing and $f(0)=1$.
Alternatively since
$$(\lambda x+(1-\lambda)y)^{\alpha}\le(\lambda x)^{\alpha}+((1-\lambda)y)^{\alpha}=\lambda^{\alpha}x^{\alpha}+(1-\lambda)^{\alpha}y^{\alpha}$$
is clear when $x=0$ we may assume $x\neq0$. We may also assume that $x=1$ since if it is true in this case then:
$(\lambda x+(1-\lambda)y)^{\alpha}=x^{\alpha}(\lambda+(1-\lambda)\frac{y}{x})^{\alpha}\le x^{\alpha}(\lambda^{\alpha}+(1-\lambda)^{\alpha}\frac{y^{\alpha}}{x^{\alpha}})=\lambda^{\alpha}x^{\alpha}+(1-\lambda)^{\alpha}y^{\alpha}$
Now we look at the function $f(x)=(1-\lambda)^{\alpha}y^{\alpha}-(\lambda+(1-\lambda)y)^{\alpha}$ for $0\le y\le1$ (since $0\le y\le x$). $f(0)=-\lambda^{\alpha}$ and $$f'(y)=\frac{(1-\lambda)^{\alpha}}{y^{1-\alpha}}-\frac{(1-\lambda)}{(\lambda+(1-\lambda)y)^{1-\alpha}}\ge\frac{(1-\lambda)^{\alpha}}{y^{1-\alpha}}-\frac{(1-\lambda)}{(1-\lambda)^{1-\alpha}y^{\alpha}}=0$$
