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No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$ --- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that $$ e(f,Q)\geq e(f,I) $$ for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve $$ \min_{f\in \mathcal{X}} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt, $$ with $\mathcal{X}=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand-> textbook question. The infimum will be in $\bar{X}$ in general, and $f$ satisfies either the Euler-Lagrange equation, or is constant. To find a simple answer, let us assume that $0<a<b$. Then $f>0$ in $\mathcal{X}$, and if you set $g=f^4$, you find that your problem is also $$ \min_{g\in \mathcal{Y}} \int_a^b \left( g + \frac{1}{16} \left(\frac{dg}{dt}\right)^2\right) \,dt, $$ with $\mathcal{Y}=\{f \in W^{1,2}(a,b)~:~g(a)=a^4,g(b)=b^4, g^\prime>0~a.e.\}$, a convex set. Now, the Euler-Lagrange equation is simply $$1+\frac{1}{8}g^{\prime\prime}=0.$$ You can solve it by hand, and playing with it, you find for example that when $$a> \frac{\sqrt{6}}{8}\frac{\left(3554+2(33)^{3/2}\right)^{1/3}}{\sqrt{7}\left(3554+2(33)^{3/2}\right)^{2/3}+(33)^{3/2}+116\left(3554+2(33)^{3/2}\right)^{1/3}+1777)},$$ the solution with $g(a)=a^4$ and $g(b)=b^4$ is strictly increasing and therefore is in $\mathcal{Y}$, whereas otherwise the positivity of the gradient constraint comes into play for suitable $b$s.

No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$ --- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that $$ e(f,Q)\geq e(f,I) $$ for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve $$ \min_{f\in \mathcal{X}} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt, $$ with $\mathcal{X}=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand-> textbook question. The infimum will be in $\bar{\mathcal{X}}$ in general, and $f$ satisfies either the Euler-Lagrange equation, or is constant. To find a simple answer, let us assume that $0<a<b$. Then $f>0$ in $\mathcal{X}$, and if you set $g=f^4$, you find that your problem is also $$ \min_{g\in \mathcal{Y}} \int_a^b \left( g + \frac{1}{16} \left(\frac{dg}{dt}\right)^2\right) \,dt, $$ with $\mathcal{Y}=\{f \in W^{1,2}(a,b)~:~g(a)=a^4,g(b)=b^4, g^\prime>0~a.e.\}$, a convex, open set. Now, the Euler-Lagrange equation is simply $$1+\frac{1}{8}g^{\prime\prime}=0.$$ You can solve it by hand, and playing with it, you find for example that when $$a> \frac{\sqrt{6}}{8}\frac{\left(3554+2(33)^{3/2}\right)^{1/3}}{\sqrt{7}\left(3554+2(33)^{3/2}\right)^{2/3}+(33)^{3/2}+116\left(3554+2(33)^{3/2}\right)^{1/3}+1777)},$$ the solution with $g(a)=a^4$ and $g(b)=b^4$ is strictly increasing and therefore is in $\mathcal{Y}$, whereas otherwise the positivity of the gradient constraint comes into play for suitable $b$s.

No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$ --- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that $$ e(f,Q)\geq e(f,I) $$ for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve $$ \min_{f\in X} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt, $$ with $X=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand..

No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$ --- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that $$ e(f,Q)\geq e(f,I) $$ for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve $$ \min_{f\in \mathcal{X}} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt, $$ with $\mathcal{X}=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand-> textbook question. The infimum will be in $\bar{X}$ in general, and $f$ satisfies either the Euler-Lagrange equation, or is constant. To find a simple answer, let us assume that $0<a<b$. Then $f>0$ in $\mathcal{X}$, and if you set $g=f^4$, you find that your problem is also $$ \min_{g\in \mathcal{Y}} \int_a^b \left( g + \frac{1}{16} \left(\frac{dg}{dt}\right)^2\right) \,dt, $$ with $\mathcal{Y}=\{f \in W^{1,2}(a,b)~:~g(a)=a^4,g(b)=b^4, g^\prime>0~a.e.\}$, a convex set. Now, the Euler-Lagrange equation is simply $$1+\frac{1}{8}g^{\prime\prime}=0.$$ You can solve it by hand, and playing with it, you find for example that when $$a> \frac{\sqrt{6}}{8}\frac{\left(3554+2(33)^{3/2}\right)^{1/3}}{\sqrt{7}\left(3554+2(33)^{3/2}\right)^{2/3}+(33)^{3/2}+116\left(3554+2(33)^{3/2}\right)^{1/3}+1777)},$$ the solution with $g(a)=a^4$ and $g(b)=b^4$ is strictly increasing and therefore is in $\mathcal{Y}$, whereas otherwise the positivity of the gradient constraint comes into play for suitable $b$s.

No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$

No. The reason is that $$e(f,Q)=\int_a^b \bigg\{f^4\bigg[1+\|\frac{d}{dr}Q\|^2+f^2\dot f^2\bigg]\bigg\} \,dr, $$ with $b\geq f\geq a$ (since $f$ is continuous and monotonous) gives $$ e(f,Q) \leq \left(1+\frac{b^4}{2}\right)\left(\|f\|^4_{W^{1,4}} + \|Q\|^2_{W^{1,2}}\right) $$
So your inequality would imply that there exists $C_1,C_2>0$ such that for all $f,Q\in\mathcal{E}$ $$ C_1\|Q\|^2_{W^{1,2}} + C_2\|f\|^4_{W^{1,4}} \geq \|Q\|^4_{W^{1,4}}. $$ Now, fix $f=a+b\frac{t-a}{b-a}$, and it becomes for all $Q\in\mathcal{E}$ $$ C_3(\|Q\|^2_{W^{1,2}} + 1) \geq \|Q\|^4_{W^{1,4}}, $$ if you cook-up a sequence $Q_n$ with $\|Q_n\|_{W^{1,2}}=1$ and $\|Q_n\|_{W^{1,4}}>n$ you have a contradiction.

Note that $W^{1,4}$ is not natural, since you can also obtain a bound from above of the form $$ e(f,Q) \leq C(a,b)\left(\|f\|^2_{W^{1,2}} + \|Q\|^2_{W^{1,2}}\right). $$ --- edited after the comment to the first part of the answer--

If you are looking for a minimizer, it is immediate to note that $$ e(f,Q)\geq e(f,I) $$ for any $f$ since it is a sum of squares, and $(f,I)\in \mathcal{E}$. Therefore you just want to solve $$ \min_{f\in X} \int_a^b \left( f^4 +f^6 \left(\frac{df}{dt}\right)^2\right) \,dt, $$ with $X=\{f \in W^{1,2}(a,b)~:~f(a)=a,f(b)=b, f^\prime>0~a.e.\}$, a convex set. A calculus of variation problem in one dimension with a monotone integrand..