For $k(s)\geq0$ and $\delta\geq0$, show that the inequality $$ \delta+\delta\int_{\tau}^{t}k(s)\exp\biggl(\int_{\tau}^{t}k(r)dr\biggl)ds\leq \delta\exp\biggl(\int_{\tau}^{t}k(s)ds\biggl)$$ holds.
If you have to know I was trying to show some Gronwall type inequality. In doing so I arrived at the above inequality.
Edit: Here is the exact question
On
If as I suspect the inner integral is between $\tau$ and $s$, the inequality is in fact an equality. Observe that $$ k(s)\exp\Bigl(\int_{\tau}^{s}k(r)dr\Bigl)=\frac{d}{ds}\exp\Bigl(\int_{\tau}^{s}k(r)\,dr\Bigr). $$
The inequalities in the Edit are difficult to follow, since there are no limits of integration, but the first inequality seems wrong to me: since $k\ge0$, $\exp(\int k(r)\,dr)\ge1$ (unless the lower limit is grater that the upper limit.)
I don't think this inequality is always true.
I don't see the purpose of $\delta$, you can divide through by it.
Let $A=\int_T^tk(x)dx$. Then your inequality becomes:
$$1+e^A\int_T^tk(s)ds\leq e^A$$
$$1+Ae^A\leq e^A$$
$$e^{-A}+A\leq1$$
which is clearly not true for all $A>0$.