I solved an integral inequality problem using geometrical methods. However, I just cannot satisfy with them and want a without-geometrical-intuition proof, and I couldn't find one.
Proof the following inequality holds when $a>0, b>0$ $$\int_{0}^{a}{\ln(x+1)\mathrm{dx}}+\int_{0}^{b}{(e^x-1)\mathrm{dx}}\geqslant ab$$
I would prefer a solution which does not involve actually solving the integral, but any proof without geometry would be a big help.
Thank you in advance!
On
As pointed out by almagest, I illegitimately disregard one term. As it is right now, this is not a proof.
Notice that for all $x\geqslant 0$, one has the following inequalities: $$\begin{align}\ln(x+1)&\geqslant x-\frac{x^2}{2}\\e^x&\geqslant x-1\end{align}.$$ Therefore, one gets: $$\int_0^a\ln(x+1)\,\mathrm{d}x+\int_0^b(e^x-1)\,\mathrm{d}x\geqslant\frac{a^2+b^2}{2}-\frac{a^3}{6}\geqslant\frac{a^2+b^2}{2}.$$ Besides, since $(a-b)^2\geqslant 0$, one has: $a^2+b^2\geqslant 2ab$. Whence, the result.
On
The function $$f(a,b):=\int_0^a\log(1+x)\>dx+\int_0^b(e^x-1)\>dx$$ grows without bounds when $a\to\infty$ or $b\to\infty$. It therefore assumes a global minimum under the constraints $$a>0,\quad b>0,\qquad ab=M$$ for given $M>0$, and this minimum will be brought to the fore by Lagrange's method. We obtain the equations $$\log(1+a)-\lambda b=0,\qquad e^b-1-\lambda a=0,$$ which immediately lead to $$a\log(1+a)=b(e^b-1)\ .\tag{1}$$ Given $a>0$ one easily guesses that $b=\log(1+a)$ solves $(1)$, and since the right hand side of $(1)$ is monotone for $b>0$ this is the only solution. The condition $ab=M$ then determines $a$ via $a\log(1+a)=M$. Call the pair of values so obtained $(a_M,b_M)$.
Now let arbitrary $a>0$, $b>0$ be given, and put $ab=:M$. Then $$f(a,b)\geq f(a_M,b_M)=\ldots=a_M\log(1+a_M)=M=ab\ .$$
Both these are straightforward integrals. The first is
$$((x+1)\ln(x+1)-x)\bracevert_0^a=(a+1)\ln(a+1)-a$$ and the second is $$(e^x-x)\bracevert_0^b=e^b-b-1$$
Let us put $b=ka$. Then we have to show that for all $a>0,k>0$ we have $$f(a,k)=(a+1)\ln(a+1)-a+e^{ka}-ka-1-ka^2\ge0$$ Differentiating wrt $k$, we get $$\frac{\partial f}{\partial k}=a(e^{ka}-1-a)$$ Putting $k_a=\frac{\ln(1+a)}{a}$, we see that $\frac{\partial f}{\partial k}$ is $<0,=0,>0$ according as for $k<k_a,k=k_a,k>k_a$. So we have a minimum at $k=k_a$. We have $$f(a,k_a)=(a+1)\ln(a+1)-a+(1+a)-\ln(1+a)-1-a\ln(1+a)=0$$ This result holds for all $a>0$ so the result is established.
$\textbf{Comment}$
At first sight this is an irritating result, because it seems really weak. The exponential will surely swamp everything else, except maybe for small $b$. Then it turns out to be much harder to prove than expected. The argument above shows that for any $a$ you can find a $b$ which gives equality and vice versa, so the result is not as weak as it looks.
For example, if we take $a=1$, then $k_a=\ln2$, so the corresponding $b=\ln2\approx0.6931$ and $\int_0^1\ln(x+1)\ dx=2\ln2-1$ and $\int_0^{\ln2}e^x-1\ dx=1-\ln2$. So their sum is $\ln2=ab$.