How to prove a function is strictly increasing?

Question

Let $f(x)$ be a continuous strictly increasing smooth function on $[0,1]$ with $f(0)=0$ and $f(1)=1$. In addition, $F(t)=\int_0^t f(x)dx$ and $\phi>1$. Prove that the following function $h(t)$ is strictly increasing with respect to $t$ on $(0,1)$, where

$$h(t)=\frac{1-t+F(t)+t(1-f(t))}{(\phi-t)(1-f(t))}-\frac{F(1)}{1-t+F(t)-F(1)+(\phi-t)(1-f(t))}.$$

The numerator of the derivative of $h(t)$ is

$$ \left((1-t+F(t))(\phi-t)f'(t)+(1-f(t))(1-t+F(t)+t(1-f(t)))\right)\left(1-t+F(t)+(\phi-t)(1-f(t))-F(1)\right)^2- F(1)((\phi-t)f'(t)+2(1-f(t)))(\phi-t)^2(1-f(t))^2, $$ and the denominator of the derivative of $h(t)$ is $$(\phi-t)^2(1-f(t))^2(1-t+F(t)+(\phi-t)(1-f(t))-F(1))^2,$$ where $f'$ is the first order derivative of $f$.

I tried to prove the numerator of the derivative is positive. To do so, I relaxed some of the terms to make it comparable. However, numerically all those relaxation methods failed. The lower bound $0$ seems to be too tight.

By the way, it can be shown both $\frac{1-t+F(t)+t(1-f(t))}{(\phi-t)(1-f(t))}$ and $\frac{F(1)}{1-t+F(t)+(\phi-t)(1-f(t))-F(1)}$ in $h$ are increasing convex functions.

Does anyone have any suggestions on how to approach this problem? Thank you so much!

Newly Edited:

I can show that $h'(t)>0$ when $\phi$ is large for certain function $f$ satisfying $1-t+F(t)>t(1-f(t))$ (It can be easily checked that all $f$ above $f(t)=t^2$ satisfies this condition). To see this, rearrange the numerator of the derivative, we get $$(\phi-t)f'(t)\bigg((1-t+F(t))(1-t+F(t)+(\phi-t)(1-f(t)))^2-F(1)(\phi-t)^2(1-f(t))^2\bigg)+2(1-f(t))\bigg(\frac{1-t+F(t)+t(1-f(t))}{2}(1-t+F(t)+(\phi-t)(1-f(t)))^2-F(1)(\phi-t)^2(1-f(t))^2\bigg).$$ The first term is positive while the second term might be not. To simplify the notation, let $A(t)=1-t+F(t)$, $A'(t)=-(1-f(t))$, and $B(t)=(\phi-t)(1-f(t))$. If the second term is negative, then we show $$(\phi-t)f'(t)\bigg(A(A+B-F(1))^2-F(1)B^2\bigg)+2(1-f(t))\bigg(\frac{A-tA'}{2}(A+B-F(1))^2-F(1)B^2\bigg)$$ is positive. It is equivalent to show $$(\phi-t)f'(t)\left(\sqrt{A}(A-F(1))+\left(\sqrt{A}-\sqrt{F(1)}\right)B\right)\left(\sqrt{A}(A-F(1))+\left(\sqrt{A}+\sqrt{F(1)}\right)B\right) >-2(1-f(t)) \left(\sqrt{\frac{A-tA'}{2}}(A-F(1))+\left(\sqrt{\frac{A-tA'}{2}}-\sqrt{F(1)}\right)B\right)\\ \left(\sqrt{\frac{A-tA'}{2}}(A-F(1)) +\left(\sqrt{\frac{A-tA'}{2}}+\sqrt{F(1)}\right)B\right).$$

Because $\sqrt{A}(A-F(1))+\left(\sqrt{A}+\sqrt{F(1)}\right)B>\sqrt{\frac{A-tA'}{2}}(A-F(1)) +\left(\sqrt{\frac{A-tA'}{2}}+\sqrt{F(1)}\right)B>0$ by the provided condition, it is enough to show $$(\phi-t)f'(t)\left(\sqrt{A}(A-F(1))+\left(\sqrt{A}-\sqrt{F(1)}\right)B\right) >-2(1-f(t)) \left(\sqrt{\frac{A-tA'}{2}}(A-F(1))+\left(\sqrt{\frac{A-tA'}{2}}-\sqrt{F(1)}\right)B\right),$$

which is equivalent to show $$\left((\phi-t)f'(t)\sqrt{A}+2(1-f(t))\sqrt{\frac{A-tA'}{2}}\right)(A-F(1))>\left(2(1-f(t))\left(\sqrt{F(1)}-\sqrt{\frac{A-tA'}{2}}\right)-(\phi-t)f'(t)\left(\sqrt{A}-\sqrt{F(1)}\right)\right)B.$$ The LHS is always positive, while on the RHS, the coefficient of $B$ is negative when $\phi$ is large enough, thus $h'(t)>0$.

Answer 1

Do you have a probabilistic interpretation of this monster? If $Y$ is a random variable with density $g(y)=f'(y)$ I found that

$$h(t)=\frac{1-\int_0^tyg(y)dy}{(\phi-t)\int^1_tg(y)dy}-\frac{1-E(Y)}{\int^1_t(y-t)g(y)dy + (\phi-t)\int^1_tg(y)dy}.$$ This suggests to check the property when $g(y)dy$ is replaced by the Dirac measure $\delta_{y_0}$ and to compute explicitely $h(t)$ in this case, namely for $t<y_0$ and for $t>y_0.$

Edit. Keeping the above idea, for $t<y_0$ and denoting for simplicity $u=\phi -t$ and $v=y_0-t$ I have found that the sign of $h'$ is the sign of

$$ u^2(3y_0-1)+uv(2-y_0)+v^2$$ which is negative if $y_0<1/3$ and $\phi$ is large. Unfortunately, I think that your conjecture is false in general.