By am-gm $$A\ge 2\left (\sqrt{\frac{1}{xy}}+\sqrt{xy}\right )=\frac{4(x+y)}{\sqrt{xy}}\ge 8$$Equality when $x=y$.

Was about to say similar to 2020 ISL A3 but then saw title. Nice one.

Let $a,b$ be positive real numbers such that $k(a+b)=1+ab. $ Prove that$$a+b+\frac{1}{a}+\frac{1}{b}\geq 4k.$$Where $k\geq 1.$ My solution: $$1+ab=k(a+b)\geq 2k\sqrt{ab}\implies ab+\frac{1}{ab}\geq4k^2-2$$$$a+b+\frac{1}{a}+\frac{1}{b}=(a+b)\left(1+\frac{1}{ab}\right)=\frac{(1+ab)^2}{kab}=\frac{1}{k}\left(ab+\frac{1}{ab}\right)+\frac{2}{k}\geq 4k.$$Equality holds when $a=b=k-\sqrt{k^2-1}$ or $a=b=k+\sqrt{k^2-1}.$ Suppose that $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+bd$. Find the smallest possible value of $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$$

sqing wrote: Suppose that $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+bd$. Find the smallest possible value of $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$$ 2020 ISL A3

parmenides51 wrote: If positive reals $x,y$ are such that $2(x+y)=1+xy$, find the minimum value of expression $$A=x+\frac{1}{x}+y+\frac{1}{y}$$ From the initial condition, $y=\frac{2x-1}{x-2}$. So we need to minimize $x+\frac{1}{x}+\frac{2x-1}{x-2}+\frac{1}{\frac{2x-1}{x-2}}$. Writing this with a common denominator, we have $\frac{2(x^2-1)^2}{(x-2)x(2x-1)}$. Let $f(x)=\frac{2(x^2-1)^2}{(x-2)x(2x-1)}$. Its first derivative is $\frac{4(x-1)^2(x^4-5x^3+6x^2-5x+1)}{(-2+x)^2x^2(-1+2x)^2}$ by the quotient rule. Roots of the derivative are $\pm1,2\pm \sqrt3$ so the minimum of $f(x)$ will occur at one of these points. Checking, the minimum of $f(x)$ is $f(2\pm\sqrt 3)=8$.

Suppose that $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+bd$. Find the smallest possible value of

Let $a=c=1,b=x,c=y,$ we have parmenides51 wrote: If positive reals $x,y$ are such that $2(x+y)=1+xy$, find the minimum value of expression $$A=x+\frac{1}{x}+y+\frac{1}{y}$$ $$1+xy=2(x+y)\geq 4\sqrt{xy}\implies xy+\frac{1}{xy}\geq 14$$$$x+\frac{1}{x}+y+\frac{1}{y}=(x+y)\left(1+\frac{1}{xy}\right)=\frac{(1+xy)^2}{2xy}=\frac{1}{2}\left(xy+\frac{1}{xy}\right)+1\geq 8.$$Equality holds when $x=y=2-\sqrt 3$ or $x=y=2+\sqrt 3.$

$A $ is just $(x+y)\frac{xy+1}{xy} = \frac{2(x+y)^2}{xy} \geqslant 8$ by AM-GM. Equality happens when $x=y$ and $x^2+1=4x$. The @above inequality follows since we can just invert both $x,y$ and the condition is kept.

We will use $AM-GM$ once on $x, y$ and once on $\frac{1}{x}, \frac{1}{y}$. * $x + y \geq 2\sqrt{xy}$ * $\frac{1}{x} + \frac{1}{y} \geq 2\sqrt{\frac{1}{xy}}$ Combining, we get: $A \geq 2(\sqrt{\frac{1}{xy}} + \sqrt{xy}) = 2\frac{xy + 1}{\sqrt{xy}} = 2 \frac{2(x + y)}{\sqrt{xy}}$ $\geq 2 \frac{2 * 2\sqrt{xy}}{\sqrt{xy}} = 2 * 4 = 8$ This means $A \geq 8$, so the smallest value of A is 8.

Nice way to use an IMO SL problem With $s = x+y$, $p=xy$, the condition is equivalent to $s = \frac{1+p}{2}$ and we wish to minimize $s + \frac{s}{p} = \frac{(p+1)^2}{2p}$. From $s\geq 2\sqrt{p}$ (with equality only when $x=y$) and the given condition we have $\frac{1+p}{2} \geq 2\sqrt{p}$, so $\frac{(p+1)^2}{2p} \geq 8$. Equality holds when $x=y$ are roots of $4t = 1+t^2$, i.e. $x=y= 2\pm \sqrt{3}$.