# From Six Variables to Four - It's All the Same

### Solution

Let $\displaystyle P=\sum_{all}a^2.\,$ By the Cauchy-Schwarz inequality,

$\displaystyle a^2+b^2\ge\frac{1}{2}(a+b)^2=\frac{t^2}{2},\;\text{with}\,2\le t\le 3,\\ \displaystyle c^2+d^2\ge\frac{1}{2}(c+d)^2=\frac{(3-t)^2}{2}\\ \displaystyle P\ge\frac{t^2+(4-t)^2}{2}=\frac{\displaystyle 2\left(t-\frac{3}{2}\right)^2+\frac{9}{2}}{2}\ge\frac{5}{2},\;\text{for}\, 2\le t\le 3.$

It follows that $\min P\displaystyle =\frac{5}{2},\,$ where $a=b=1,\,$ $\displaystyle c=d=\frac{1}{2}.$

Further,

(*)

$P =(a+b)^2-2ab+(c+d)^2-2cd\le t^2+(3-t)^2-2ab.$

Since $a,b\ge 1,\,$ we also have $(a-1)(b-1)\ge 0,$

implying,

(**)

$ab\ge a+b-1.$

From (*) and (**) it follows that

\begin{align} P&\le (3-t)^2+t^2-2t+2=2t^2-8t+11\\ &=2(t-2)^2+3\le 5,\;\text{for}\, 2\le t\le 3. \end{align}

Thus, $\max P=5,\,$ where $t=3,\,$ i.e., $a=2,\,$ $b=1,\,$ $c=d=0.$

### Acknowledgment

I am grateful to Leo Giugiuc who has kindly posted the problem of his at the CutTheKnotMath facebook page. He later posted a similar problem with six variables. The above is an application of Richdad Phuc's solution of that problem to the current, simpler one. Worked like a charm.

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