# A conjectural inequality of the form $\int_a^b g(B'_1(t)) dt \le \int_a^b g(B'_2(t)) dt$ with convex increasing $g$

Assume $g:[0,\infty)\to [0,\infty)$ be strictly convex and increasing monotonic function and $B_1:[0,\infty)\to [0,\infty)$ be convex and increasing monotonic function and $B_2:[0,\infty)\to [0,\infty)$ is increasing monotonic function and $B_1(t)>B_2(t)$ for every $a<t<b$ and $B_1(t)=B_2(t)$ for $t=a$ , $t=b$ then I guess that we can show $$\int_a^b g(B'_1(t)) dt \le \int_a^b g(B'_2(t)) dt$$

I have tried to use Jensen's inequality or other similar inequalities and also I have tried to use calculus of variation methods but I have not reached to any good place.

Can somebody give a counterexample or give hint to proof this even with some modification on hypothesizes

• Integrate by parts. – ec92 Dec 16 '14 at 20:28
• @ec92 could you elaborate? – TZakrevskiy Dec 16 '14 at 20:32
• Or, even simpler, make the area over the top graph of stone, the area over the bottom one of paper, take scissors, and start cutting pieces off the area over the bottom graph carefully avoiding the stone. Translate what you see into induction/limits/whatever other math. dialect you like. – fedja Dec 17 '14 at 4:19
• @fedja : thanks .I have reached this idea myself, but I didn't consider it as a proof. I have guessed this inequality exactly in this way that you explained. – Finish Dec 17 '14 at 10:10
• It is a finite process when $B_1$ is piecewise linear, so it is completely rigorous in that case. The general case follows by approximation as soon as you show that the piecewise linear functions inscribed into the graph of $B_1$ converge to $B_1$ in a strong enough sense to justify passing to the limit in the integral. In general, "a proof" is not as much a formal argument as a clear mental picture translated into a comprehensible language, and the ability to translate just comes with training and experience. :-) – fedja Dec 17 '14 at 11:10