11 years ago, I asked this question. Now I have a new question:

Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that $$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$ for every $t$ in $[0,a]$. Prove or disprove the following conjecture: $$\dfrac{a^3}{12}\ge\int_{0}^{a}|F(x)-\frac{1}{2}x|^2dx$$

with equality if and only if $F(x)=x$.

My attemp: Using macavity methods it seems we cannot solve this problem.

We need show $$\int_{0}^{a}F^2(x)\le\int_{0}^{a}xF(x)dx$$

if we use the same methods.

$$G_t = \int_0^t F dx \le \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot \left( \int_0^t 1 dx \right)^{\frac{2}{3}} = \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot t^{\frac{2}{3}} \le {G_t}^{\frac{2}{3}}\cdot t^{\frac{2}{3}} \quad \implies \sqrt{G_t} \le t, $$ so $G_{t}\le t^2$. I think this step can't following, because we must have $G_{t}\le\dfrac{t^2}{2}$, since $F(x)=x$, so $G_{t}=\dfrac{t^2}{2}$.

11 years ago, I asked this question. Now I have a new question:

Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that $$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$ for every $t$ in $[0,a]$. Prove or disprove the following conjecture: $$\dfrac{a^3}{12}\ge\int_{0}^{a}|F(x)-\frac{1}{2}x|^2dx$$

My attemp: Using macavity methods it seems we cannot solve this problem.

We need show $$\int_{0}^{a}F^2(x)\le\int_{0}^{a}xF(x)dx$$

if we use the same methods.

$$G_t = \int_0^t F dx \le \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot \left( \int_0^t 1 dx \right)^{\frac{2}{3}} = \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot t^{\frac{2}{3}} \le {G_t}^{\frac{2}{3}}\cdot t^{\frac{2}{3}} \quad \implies \sqrt{G_t} \le t, $$ so $G_{t}\le t^2$. I think this step can't following, because we must have $G_{t}\le\dfrac{t^2}{2}$, since $F(x)=x$, so $G_{t}=\dfrac{t^2}{2}$.

11 years ago, I have ask this question:1,Now I have new question  :

Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that $$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$ for every $t$ in $[0,a]$,prove or disprove the conjecture: $$\dfrac{a^3}{12}\ge\int_{0}^{a}|F(x)-\frac{1}{2}x|^2dx$$

if and only if:when $F(x)=x$

my attemp  :use Macavity methods it seem can't solve this problem

we need show $$\int_{0}^{a}F^2(x)\le\int_{0}^{a}xF(x)dx$$

if we use same methods.

$$G_t = \int_0^t F dx \le \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot \left( \int_0^t 1 dx \right)^{\frac{2}{3}} = \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot t^{\frac{2}{3}} \le {G_t}^{\frac{2}{3}}\cdot t^{\frac{2}{3}} \quad \implies \sqrt{G_t} \le t $$ ,so $G_{t}\le t^2$,I think this step can't following  ,because we must $G_{t}\le\dfrac{t^2}{2}$,since $F(x)=x$,so $G_{t}=\dfrac{t^2}{2}$

11 years ago, I asked this question. Now I have a new question:

Let $F(x)$ be nonnegative and integrable on $[0,a]$ and such that $$\left(\int_{0}^{t}F(x)dx\right)^2\ge\int_{0}^{t}F^3(x)dx$$ for every $t$ in $[0,a]$. Prove or disprove the following conjecture: $$\dfrac{a^3}{12}\ge\int_{0}^{a}|F(x)-\frac{1}{2}x|^2dx$$

with equality if and only if $F(x)=x$.

My attemp: Using macavity methods it seems we cannot solve this problem.

We need show $$\int_{0}^{a}F^2(x)\le\int_{0}^{a}xF(x)dx$$

if we use the same methods.

$$G_t = \int_0^t F dx \le \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot \left( \int_0^t 1 dx \right)^{\frac{2}{3}} = \left( \int_0^t F^{3} dx \right)^{\frac{1}{3}} \cdot t^{\frac{2}{3}} \le {G_t}^{\frac{2}{3}}\cdot t^{\frac{2}{3}} \quad \implies \sqrt{G_t} \le t, $$ so $G_{t}\le t^2$. I think this step can't following, because we must have $G_{t}\le\dfrac{t^2}{2}$, since $F(x)=x$, so $G_{t}=\dfrac{t^2}{2}$.