Show geometrically or algebraically $(\sqrt2-1)a+(\sqrt3-1)b<c$ Pythagoras 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**

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