65th Czech and Slovak Mathematical Olympiad

To prove the inequality $ab\ge 4$ we substitute from the equalities. We so obtain an estimate $$ab=\left(c+\frac{1}{d}\right)\left(d+\frac{1}{c}\right)=cd+1+1+\frac{1}{cd}\ge 4,$$ where we use in the last inequality well-known fact that $x+1/x\ge 2$ holds for all positive reals $x=cd>0$.

To find the minimum we use similar way. Substitution for $a$ and $b$ yields $$ab+cd=\left(2+cd+\frac{1}{cd}\right)+cd=2+2cd+\frac{1}{cd}.$$ Now we use an inequality $x+y\ge 2\sqrt{xy}$ which holds true for any non-negative reals $x,y$. The choice $x=2cd$, $y=1/cd$ follows $$2cd+\frac{1}{cd}\ge 2\sqrt{2}.$$ Now we see that $ab+cd\ge 2(1+\sqrt{2})$. To prove that it is the desired minimum we find some $a$, $b$, $c$, $d$ such that they makes an equality in the inequality. The equality comes in the use inequality if and only if $x=y$, it is $2cd=1/cd$. It is true e.g. for $c=1$, $d=\sqrt{2}/2$ and for that values we find $a=1+\sqrt{2}$, $b=1+\sqrt{2}/2$. Such quadruple satisfies the desired equalities and it holds $ab+cd=2(1+\sqrt{2})$ too.