<s>[INST] Show geometrically or algebraically $(\sqrt2-1)a+(\sqrt3-1)bPythagoras theorem
$$a^2+b^2=c^2$$
Show geometrically (Addressing to @blue a  Trigonographer)
(1)$$(\sqrt2-1)a+(\sqrt3-1)b<c$$
Or algebraically (general users)

(@BLue the Trigonographer)
Expand (1) 
$a\sqrt2+b\sqrt3<a+b+c$
Letting P=a+b+c be the perimeter of triangle ABC and where $A_1=a\sqrt2$, $A_2=b\sqrt3$ are two areas.
We can say $A_1+A_2<P$
I just wonderly can it be construct geometrically to show this inequality.
I have seem on his site, an amazing diagrams and beautiful proof via diagrams
 [/INST] Using Schwarz inequality we get $$\left(\sqrt 2 - 1\right)a + \left(\sqrt 3 - 1\right)b \le \sqrt{\left(\left(\sqrt 2 - 1\right)^2 + \left(\sqrt 3 - 1\right)^2\right)\left(a^2+b^2\right)} = \sqrt{7 - 2\sqrt 2 - 2 \sqrt 3}c.$$
Therefore it is enough to check that $7 - 2\sqrt 2 - 2\sqrt 3 < 1$ which I leave as an exercise.
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