By using the fact that $f(0)=1$ and by the concavity of $f$ we can say, $x^{a} f(x)+1-x^{a}=x^{a} f(x)+\left(1-x^{a}\right) f(0) \leq f\left(x^{a} \cdot x+\left(1-x^{a}\right) \cdot 0\right)=f\left(x^{a+1}\right)$ so we get $x^{a} f(x)+(1-x^{a})\leq f\left(x^{a+1}\right)$ multiplying both side by $(a+1)x^a$ we get , $(a+1) x^{2 a} f(x)+(a+1)\left(x^{a}-x^{2 a}\right) \leq(a+1) x^{a} f\left(x^{a+1}\right)$ now Integrate both side from 0 to 1 $(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x+(a+1) \int_{0}^{1}\left(x^{a}-x^{2 a}\right) \mathrm{d} x \leq(a+1) \int_{0}^{1} x^{a} f\left(x^{a+1}\right) \mathrm{d} x$ that is , $(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x+\frac{a}{2 a+1} \leq \int_{0}^{1} f(x) \mathrm{d} x$ using the fact , $(\int_0^1f(x)-\frac{1}{4})^2\geq 0$ we get . $\int_{0}^{1} f(x) \mathrm{d} x \leq\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2}+\frac{1}{4},$ pluging this value in our previous inequality and arranging a bit we get $\quad\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2}-(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x \geq\frac{a}{2 a+1}-\frac{1}{4}=\frac{2 a-1}{8 a+4}$

In order for the above solution to be complete we also need an example of function for which the equality holds. An example is $1-x.$