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
 A: Algebraic (general user).
We have $(4a-3b)^2\ge0$, so expanding $16a^2-24ab+9b^2\ge0$. Hence $25(a^2+b^2)\ge(9a^2+24ab+16b^2)$ or $5c\ge 3a+4b$.
Hence $8a+9b\le5(a+b+c)$ or $1.6a+1.8b\le a+b+c$. Since $1.6>\sqrt2$ and $1.8>\sqrt3$ this implies $a\sqrt2+b\sqrt3<a+b+c$.
A: 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.
A: $a\sqrt2+b\sqrt3<a+b+c$
$a\sqrt2+b\sqrt3<c\sin(\theta)+c\cos(\theta)+c$
$\frac{a}{c}\sqrt2+\frac{b}{c}\sqrt3<\sin(\theta)+\cos\theta)+1$
$\sqrt2\sin(\theta)+\sqrt3\cos(\theta)<\sin(\theta)+\cos(\theta)+1$
$(\sqrt2-1)\sin(\theta)+(\sqrt3-1)\cos(\theta)<1$
The maximum value of $\sin(\theta)+\cos(\theta)=\sqrt2$ at $\theta=45^0$
$(\sqrt2-1)\sin(45)+(\sqrt3-1)\cos(45)<1$
$\frac{\sqrt6-2\sqrt2+2}{2}<1$
$0.81053\cdots<1$
