Inequality:$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$

Solution 1:

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.