Inequality:$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$ For $ a,b,c,d\geq 0 $ with $ a+b = c+d = 2 $, how to  prove that $$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$$
 A: We will make repeated use of the identity $(p^2+q^2)(r^2+s^2)=(pr-qs)^2+(ps+qr)^2$.
Applying it to the first two terms yields
$$
(a^2+c^2)(a^2+d^2)=(a^2-cd)^2+(a(c+d))^2=(a^2-cd)^2+(2a)^2 \, .
$$
Similarly, the third and fourth terms multiply to $(b^2-cd)^2+(2b)^2$.
Applying the identity again to these two quantities, we have
$$
\left[(a^2-cd)^2+(2a)^2\right]\left[(b^2-cd)^2+(2b)^2\right] \\
= \left[(a^2-cd)(b^2-cd)-4ab\right]^2+\left[(a^2-cd)(2b)+(b^2-cd)(2a)\right]^2 \\
=: Q^2+R^2 \, ,
$$
and we'll now work on simplifying $Q$ and $R$ separately.
First,
\begin{eqnarray}
Q &=& (a^2-cd)(b^2-cd)-4ab \\
&=& a^2b^2-(a^2+b^2)cd+c^2d^2-4ab \\
&=& a^2b^2-(a^2+2ab+b^2)cd+2abcd + c^2d^2-4ab \\
&=& a^2b^2-4cd+2abcd+c^2d^2-4ab \\
&=& ab(ab+cd-4)+cd(ab+cd-4)\\
&=& (ab+cd)(ab+cd-4) \, .
\end{eqnarray}
Next,
\begin{eqnarray}
R &=& (a^2-cd)(2b)+(b^2-cd)(2a) \\
&=& 2ba^2-2bcd+2ab^2-2acd \\
&=& 2ab(a+b)-2cd(a+b)\\
&=&4(ab-cd) \, .
\end{eqnarray}
So the quantity we're trying to bound is equal to
$$
\left[(ab+cd)(ab+cd-4)\right]^2+[4(ab-cd)]^2 \, .
$$
Let $x=ab$, $y=cd$. Then the constraints on $a,b,c,d$ imply that $x,y$ each vary freely over $[0,1]$. So, in order to prove the inequality, it suffices to show that
$$
f(x,y)=(x+y)^2(x+y-4)^2+16(x-y)^2
$$
is bounded above by 25 on the unit square.
To show this, we first make the temporary change of variables $s=x+y$, $t=x-y$; let $F(s,t)=s^2(s-4)^2+16t^2$ be the pullback of $f$ under this coordinate change. Then $F_s=2s(s-4)(2s-4)$ is nonvanishing for $0<s<2$. It follows that $F$ has no critical points on this strip, and thus that $f$ has no critical points on the interior of the unit square. So $f$ will attain its maximum somewhere on the boundary.
Since $f$ is symmetric in its arguments, it suffices to examine the functions $f_0(y)=f(0,y)$ and $f_1(x)=f(x,1)$. We have
$$f_0(y)=y^2(y-4)^2+16y^2=y^4-8y^3+32y^2 \\
f_0'(y)=4y^3-24y^2+64y=4y(y^2-6y+16)=4y((y-3)^2+7) \, ;
$$
thus $f_0'$ is nonzero on $(0,1)$, so $f_0$ has no critical points on $(0,1)$. Also,
$$
f_1(x)=(x+1)^2(x-3)^2+16(x-1)^2=((x-1)^2+1)^2\\
f_1'(x)=4(x-1)((x-2)^2+1)^2 \, ;
$$
thus $f_1'$ is also nonzero on $(0,1)$, and $f_1$ has no relevant critical points. So $f$ is maximized at one of the corners of the square.
Finally,
$$
f(0,0)=16 \\
f(0,1)=25 \\
f(1,1)=16
$$
and so the maximum value of $f$ is 25, as desired.
