Vlados021 wrote: Find all positive numbers $a$, $b$, $c$, $d$ such that $a+b+c+d=1$ and $$ \max\left\{\frac{a^2}{b},\frac{b^2}{a}\right\}\cdot \min\left\{\frac{c^2}{d},\frac{d^2}{c}\right\}=(\min\{a+b,c+d\})^4. $$ Swapping $a,b$ leaves unchanged the equation, and so WLOG $a\ge b$ Swapping $c,d$ leaves unchanged the equation, and so WLOG $c\ge d$ Let $x=a+b\in(0,1)$ Let $u=\frac ab\ge 1$ and $v=\frac cd\ge 1$ So $(a,b,c,d)=\left(\frac {ux}{u+1},\frac x{u+1},\frac{v(1-x)}{v+1},\frac{1-x}{v+1}\right)$ Equation becomes $\frac{u^2}{(u+1)v(v+1)}x(1-x)=\min(x,1-x)^4$ So $v(v+1)=\frac{u^2x(1-x)}{(u+1)\min(x,1-x)^4}$ Considering this as an equation in $v$, we always have a positive solution. In order this positive solution be $\ge 1$, we have the constraint $\frac{u^2}{u+1}\ge \frac{2\min(x,1-x)^4}{x(1-x)}$ And so the infinite set of all solutions : Choose any $x\in(0,1)$ Choose any $u\ge 1$ such that $\frac{u^2}{u+1}\ge \frac{2\min(x,1-x)^4}{x(1-x)}$ Choose $v$ as the positive root of equation $v(v+1)=\frac{u^2x(1-x)}{(u+1)\min(x,1-x)^4}$ And then you have the solution $\boxed{(a,b,c,d)=\left(\frac {ux}{u+1},\frac x{u+1},\frac{v(1-x)}{v+1},\frac{1-x}{v+1}\right)}$ Plus the solutions obtained swapping $a,b$ and/or swapping c,d

Vlados021 wrote: Find all positive numbers $a$, $b$, $c$, $d$ such that $a+b+c+d=1$ and $$ \max\left\{\frac{a^2}{b},\frac{b^2}{a}\right\}\cdot \min\left\{\frac{c^2}{d},\frac{d^2}{c}\right\}=(\min\{a+b,c+d\})^4. $$ I think that there should be $$\max\left\{\frac{a^2}{b},\frac{b^2}{a}\right\}\cdot \max\left\{\frac{c^2}{d},\frac{d^2}{c}\right\}=(\min\{a+b,c+d\})^4. $$