For any $0 \leq x \leq y \leq 1$, define $f(y;x):=\frac{y^2}{2}-\frac{2 y^3}{3}+\frac{y^4}{4} - \frac{x^2}{2} + \frac{x^3}{3}$ and $g(y;x):=\frac{y^2}{3}-\frac{2 y^3}{4}+\frac{y^4}{5} - \frac{x^2}{3} + \frac{x^3}{4}$. Show that for any $x' \in [0,1]$, if there exists $y' \in [0,1]$ such that $f(y';x')=0$, then $g(y';x') \leq 0$.
It is straightforward to "verify" it numerically by matlab, and I have not find any counter examples. I would really appreciate if anyone can help out.
(P.S.: In fact, I have numerically "verified" that a more general statement holds: For any $0 \leq x \leq y \leq 1$ and $n \in \mathbb{Z}_{++}$, define $f_n(y;x):=y^2\sum_{k=0}^{n} \frac{(-1)^k y^k C_{n}^k}{k+2}- x^2\sum_{k=0}^{n-1} \frac{(-1)^k x^k C_{n-1}^k}{k+2}$ and $g_n(y;x):=y^2\sum_{k=0}^{n} \frac{(-1)^k y^k C_{n}^k}{k+3}- x^2\sum_{k=0}^{n-1} \frac{(-1)^k x^k C_{n-1}^k}{k+3}$; for any $x' \in [0,1]$, if there exists $y' \in [0,1]$ such that $f_n(y';x')=0$, then $g_n(y';x') \leq 0$. But I guess the special case of this more general problem can help gain some intuition)
Let $h_1(t)=t(1-t)^2$ and $h_2(t)=t(1-t)$, so $$ f(y;x)=\int_0^yh_1(t)\;dt-\int_0^xh_2(t)\;dt, $$ $$ g(y;x)=\frac1y\int_0^yth_1(t)\;dt-\frac1x\int_0^xth_2(t)\;dt. $$ Note that $0<h_1(t)<h_2(t)$ for $t\in(0,1)$, and if $\lambda\geq1$ then $$ \frac{h_1(\lambda t)}{h_2(t)} =\lambda\left(\lambda-\frac{\lambda-1}{1-t}\right)(1-\lambda t) $$ is a decreasing function on $(0,\frac1\lambda)$. In particular if $\lambda\mu<1$ then $$ \frac{h_1(\lambda t)}{h_1(\lambda\mu)}>\frac{h_2(t)}{h_2(\mu)} $$ for $t\in(0,\mu)$. Integrating, $$ \frac{F_1(\lambda\mu)}{\lambda\mu h_1(\lambda\mu)}> \frac{F_2(\mu)}{\mu h_2(\mu)} $$ where $$ F_1(y)=\int_0^yh_1(t)\;dt,\hspace{10mm} F_2(x)=\int_0^xh_2(t)\;dt. $$ Note that $F_1$ and $F_2$ are increasing on $[0,1]$, and $F_1(t)\leq F_2(t)$. For any $a\in(0,F_1(1)]$ we have $F_1^{-1}(a)\geq F_2^{-1}(a)$, so putting $\mu=F_2^{-1}(a)$ and $\lambda=F_1^{-1}(a)/F_2^{-1}(a)$, $$ F_2^{-1}(a)h_2(F_2^{-1}(a))>F_1^{-1}(a)h_1(F_1^{-1}(a)). $$ Thus $$ \frac{d}{da}\left(\ln\left(\frac{F_2^{-1}(a)}{F_1^{-1}(a)}\right)\right) =\frac1{F_2^{-1}(a)h_2(F_2^{-1}(a))}-\frac1{F_1^{-1}(a)h_1(F_1^{-1}(a))}<0, $$ so for $b\in(0,a)$ we have $$ \frac{F_2^{-1}(b)}{F_1^{-1}(b)}>\frac{F_2^{-1}(a)}{F_1^{-1}(a)}. $$ Suppose $f(y';\,x')=0$, and put $a=F_1(y')=F_2(x')$. Then $$ \frac1{x'}F_2^{-1}(b)>\frac1{y'}F_1^{-1}(b). $$ Integrating, $$ \frac1{x'}\int_0^aF_2^{-1}(b)\;db\geq\frac1{y'}\int_0^aF_1^{-1}(b)\;db. $$ The change of variable $b=F_2(t)$ gives $$ \int_0^aF_2^{-1}(b)\;db=\int_0^{x'}th_2(t)\;dt $$ and similarly for the RHS, so $g(y';\,x')\leq0$.