\section*{Problem 15}

What is the minimal value of \(\mathrm{b} /(\mathrm{c}+\mathrm{d})+\mathrm{c} /(\mathrm{a}+\mathrm{b})\) for positive real numbers \(\mathrm{b}\) and \(\mathrm{c}\) and nonnegative real numbers a and \(\mathrm{d}\) such that \(\mathrm{b}+\mathrm{c} \geq \mathrm{a}+\mathrm{d}\) ?


\section*{Solution}

Answer: \(\sqrt{ } 2-1 / 2\).

Obviously \(\mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}\) at the minimum value, because increasing a or \(\mathrm{d}\) reduces the value. So we may take \(\mathrm{d}=\mathrm{b}+\mathrm{c}-\mathrm{a}\). We also take \(\mathrm{b}>=\mathrm{c}\) (interchanging \(\mathrm{b}\) and \(\mathrm{c}\) if necessary). Dividing through by \(\mathrm{b} / 2\) shows that there is no loss of generality in taking \(\mathrm{b}=2\), so \(0=0\), so \(\mathrm{c}^{2}+\mathrm{c}(2 \mathrm{k}-2 \mathrm{~h})+\mathrm{h}^{2}-2 \mathrm{hk}\) \(+k^{2}>=0\). Hence \(c^{2}+c k+h^{2}>=(2 h-k)(c+k)\). Hence \(c / h^{2}+1 /(c+k)>=2 / h-k / h^{2}\) with equality iff \(c=h-k\). Applying this to \(c /(a+2)+1 /(c+1-a / 2)\) where \(h=\sqrt{ }(a+2), k=1-a / 2\), we find that \(c /(a+2)+2 /(2 c+2-a)>=2 / \sqrt{(a+2)}+(a-2) /(2 a+4)\).

The allowed range for \(\mathrm{c}\) is \(0 \leq c \leq 2\) and \(0 \leq a \leq c+2\), hence \(0 \leq a \leq 4\). Put \(x=1 / \sqrt{(a+2)}\), so \(1 / \sqrt{6}<=x<=1 / \sqrt{2}\). Then \(2 / \sqrt{(} a+2)+(a-2) /(2 a+4)=2 x+1 / 2-2 x^{2}=1-(2 x-1)^{2} / 2\). We have \(0.184=(2 / \sqrt{6}-1) \leq 2 x-1 \leq \sqrt{ } 2-1=0.414\). Hence \(c /(a+2)+2 /(2 c+2-a) \geq 1-(\sqrt{ } 2-1)^{2} / 2=\sqrt{2}-\) \(1 / 2\).

We can easily check that the minimum is achieved at \(\mathrm{b}=2, \mathrm{c}=\sqrt{2}-1, \mathrm{a}=0, \mathrm{~d}=\sqrt{2}+1\).

