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The structure functions $F_2^{\rm ep}$ and $F_2^{\rm en}$ for electron-proton and electron-neutron deep inelastic scattering deduced from the model of valence quarks and sea quarks of proton and neutron have the expressions$$ \frac{1}{x}F_2^{\rm ep}(x)=\Big(\frac{2}{3}\Big)^2[u_p(x)+\bar{u}_p(x)] +\Big(\frac{1}{3}\Big)^2[d_p(x)+\bar{d}_p(x)] +\Big(\frac{1}{3}\Big)^2[s_p(x)+\bar{s}_p(x)],\\ \frac{1}{x}F_2^{\rm en}(x)=\Big(\frac{2}{3}\Big)^2[u_n(x)+\bar{u}_n(x)] +\Big(\frac{1}{3}\Big)^2[d_n(x)+\bar{d}_n(x)] +\Big(\frac{1}{3}\Big)^2[s_n(x)+\bar{s}_n(x)]$$ where $u_p, d_p, s_p$ are the momentum distribution functions $f_i(x)$ for $u, d$ and $s$ quarks while $\bar{u}_p, \bar{d}_p, \bar{s}_p$ are those of $\bar{u},\bar{d}$ and $\bar{s}$ antiquarks inside the proton, respectively. If isospin symmetry is considered to be exact (very small violation), the $u$-qurks distribution in proton should be same as the $d$ quarks distribution in th neutron and vice-versa. Also, the s-quarks and $\bar{s}$-antiquarks distributions will be same in the proton and the neutron. Therefore, $$ u_p(x) =d_n(x)\equiv u(x), ~~ d_p(x) =u_n(x)\equiv d(x),~~s_p(x) =s_n(x) \equiv s(x).$$ With this simplification, the expressions for the structure cosntants become, $$ \frac{1}{x}F_2^{\rm ep}(x) = \Big(\frac{2}{3}\Big)^2[u(x)+\bar{u}_p(x)] +\Big(\frac{1}{3}\Big)^2[d(x)+\bar{d}_p(x)] +\Big(\frac{1}{3}\Big)^2[s(x)+\bar{s}(x)],\\ \frac{1}{x}F_2^{\rm en}(x) = \Big(\frac{2}{3}\Big)^2[d(x)+\bar{u}_n(x)] +\Big(\frac{1}{3}\Big)^2[u(x)+\bar{d}_n(x)] +\Big(\frac{1}{3}\Big)^2[s(x)+\bar{s}(x)].$$ Using $$u(x)=u_v(x)+u_{\rm sea}(x), d(x)=d_v(x)+d_{\rm sea}(x)$$ and realising that for other quarks and antiquarks, the contribution entirely come from sea quarks, one finally ends up getting $$ \frac{1}{x}F_2^{\rm ep}(x) = \Big(\frac{1}{9}\Big)[4u_v+d_v] +\Big(\frac{4}{3}\Big)S(x), \\ \frac{1}{x}F_2^{\rm en}(x) = \Big(\frac{1}{9}\Big)[u_v+4d_v] +\Big(\frac{4}{3}\Big)S(x).$$

In an exercise of Halzen and Martin's Particle Physics text, they've asked to show that $$\frac{1}{4}\leq\frac{F_2^{\rm en}(x)}{F_2^{\rm ep}(x)}\leq 4$$ for all values of $x$. It is mentioned that the upper limit will be realized if only $u$ quarks were present and the lower limit will be realized if only $d$ quarks are present in the proton. But how can one derive this inequality with nothing said about the neutron? They say that this is obvious (from the last two relations) but I cannot prove this result. It doesn't look obvious to me at all. Any help will be very much appreciated.

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While proving the inequality, we can ignore the $x$ depedence of the variables, since the inequality must be satisfied for any $x$. We want to show that: $$\frac{1}{4}\leq\frac{F_2^{\rm en}}{F_2^{\rm ep}}\leq 4$$ This is equivalent to showing that

  1. $f(s,u,d)=F_2^{\rm en} - 4 F_2^{\rm ep} \leq 0$
  2. $g(s,u,d)=F_2^{\rm en} - \frac14 F_2^{\rm ep} \geq 0$

where $s,u,d$ are three independent variables. Explicitly, $$f(s,u,d)=-4s -\frac53 u,\,\,\,\,\,\,\,\,\,\,\,\,g(s,u,d)=s +\frac{5}{12} d$$ These are hyperplanes which have no local minima. Therefore their absolute maximum/minimum will be found on the boundary. Since $s,u,d$ come from distribution functions, $0 \leq s,u,d \leq 1$.

The rest is simply checking each of the four boundary pieces in the two cases. However, since $0 \leq s,u,d \leq 1$, it is actually quite clear that $f$ is always negative except when $s=u=0$, where it achieves a maximum of $0$. Similarly $g$ is always positive and its minimum is zero at $s=d=0$.

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