Elegant proof of convexity

Question

I am seeking an elegant way to prove the concavity with respect to $x$ of $$f(x,t,a):=\Big(1-\frac1{(1+x)^t}\Big)\Big(1-\frac ax\Big), \quad x\ge0,\; t>0,\; a>0.$$

I put up my mundane proof as an answer below.

Answer 1

I have found an elegant proof:

$$g(x,t):=\frac1x\Big(1-\frac1{(1+x)^t}\Big)=t\int_0^1\frac1{(1+\zeta x)^{t+1}}\,d\zeta$$ is convex with respect to $x$ since the integrand is. So $-ag(x,t)$ is concave. As $h(x,t):=1-\frac1{(1+x)^t}$ is concave also with respect to $x$, so is $f(x,t,a)=h(x,t)-ag(x,t)$.

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Here is the proposition I used in the above proof.

Proposition: Function $h(x,t)$ is convex in $x$ and integrable on $t\in[0,1]$. $I(x):=\int_0^1h(x,t)dt$ is convex.

Proof: $\forall \lambda\in[0,1]$, \begin{align} I(\lambda x+(1-\lambda)y)&=\int_0^1h\big(\lambda x+(1-\lambda)y,t\big)\,dt \\ &\le \int_0^1\big(\lambda h(x,t)+(1-\lambda)h(y,t)\big)\,dt \\ &=\lambda\int_0^1h(x,t)dt+(1-\lambda)\int_0^1h(y,t)\,dt \\ &= \lambda I(x)+(1-\lambda)I(y). \end{align} So $I$ is convex.