# Minimizing $(x+y)(z+t)(\frac{a}{x} + \frac{c}{z})(\frac{b}{y} + \frac{d}{t}).$

Let $$a,b,c,d$$ be positive constants and $$x,y,z,t$$ be positive variables. Find the minimum value of $$$$(x+y)(z+t)\left(\frac{a}{x} + \frac{c}{z}\right)\left(\frac{b}{y} + \frac{d}{t}\right).$$$$

I came up with this nice problem recently but haven't found a solution yet, so I hope somebody here could help.

Using AM-GM one can easily obtain a lower bound of $$16\sqrt{abcd}$$, which is not necessarily the minimum value unless $$ad = bc$$.

• Hint: If $m,n > 0$ are constants, when does the minimum of $( x + m) ( \frac{1}{x} + n)$ occur? Dec 8, 2021 at 18:05

Hint: Use Cauchy-Schwarz (CS) inequality (thrice?)

$$(x+y)(t+z) \cdot \left(\frac ax+\frac cz\right)\left(\frac dt+\frac by \right) \geqslant (\sqrt{xt}+\sqrt{yz})^2\cdot\left(\sqrt{\frac{ad}{xt}}+\sqrt{\frac{bc}{yz}} \right)^2 \geqslant (\sqrt[4]{ad}+\sqrt[4]{bc})^4$$

P.S. Equality is when all the CS used can be equalities, i.e. the following conditions hold $$xz=yt, abtz = cdxy, (xt)^2bc =(yz)^2 ad$$.

• This is not a hint but rather an entire solution that is very nice! Thanks.
– f10w
Dec 8, 2021 at 21:07
• Extension to 3x3 looks much more difficult: math.stackexchange.com/q/4329363/31498.
– f10w
Dec 10, 2021 at 16:10

Using algebra $$F=(x+y)(z+t)\left(\frac{a}{x} + \frac{c}{z}\right)\left(\frac{b}{y} + \frac{d}{t}\right)$$ Compute the partial derivatives $$\frac{\partial F}{\partial x}=\frac{(t+z) (b t+d y) }{t x^2 y z}\color{red}{\left(c x^2-a y z\right)}=0$$ $$\frac{\partial F}{\partial y}=\frac{(t+z) (a z+c x)}{t x y^2 z} \color{red}{\left(d y^2-b t x\right)}=0$$ $$\frac{\partial F}{\partial z}=\frac{(x+y) (b t+d y) }{t x y z^2}\color{red}{\left(a z^2-c t x\right)}=0$$ $$\frac{\partial F}{\partial t}=\frac{(x+y) (a z+c x)}{t^2 x y z} \color{red}{\left(b t^2-d y z\right)}=0$$

Since all variables and parameters are positive, you need to solve for $$(x,y,z,t)$$ the "red" equations. Using the first, second and third, we have $$y=\sqrt[4]{\frac{b c}{a d}}x \qquad z=\sqrt[4]{\frac{c^3 d}{a^3 b}}x\qquad t=\sqrt[2]{\frac{c d}{a b}}x$$ Plugged in the fourth equation, we just find the interesting $$0=0$$ result.

All of the above plugged in $$F$$ gives as a result $$\Big[\sqrt[4]{ad}+\sqrt[4]{bc}\Big]^4$$

• Thanks. That's indeed a simple solution.
– f10w
Dec 9, 2021 at 19:59
• It's not trivial for the 3x3 extension though: math.stackexchange.com/q/4329363/31498
– f10w
Dec 10, 2021 at 16:10