Can the Maximum Sum of Diagonals in a Rhombus Be Proven to Be 14?

[Albert1]

A rhombus with side length 5

if diagonal BD $\geq 6$

and diagonal AC $\leq 6$

Prove : max (AC+BD)=14

 

[Albert1]

Re: Prove max(AC+BD)=14

A rhombus with side length 5

if diagonal BD $\geq 6$

and diagonal AC $\leq 6$

Prove : max (AC+BD)=14

let $BD=x=6+a,AC=y=6-b$
for $x^2+y^2=100$
$we\,\, have\,\, (a>0 ,b\leq0$)
$\therefore (6+a)^2+(6-b)^2=100$
$(a-b)^2+12(a-b)+2ab-28=0$
$a-b=-6\pm\sqrt{64-2ab}$
$\therefore max(a-b)=2 \,\, (here \,\, b=0,a=2)$
$and\,\,we\,\,get :max(y+x)=max(AC+BD)=(6-0)+(6+2)=14$