Lol. Let $a=x^2,b=y^2,c=z^2$. Then $a^2<16bc$ implies that $$x^4<16y^2z^2\iff x<2\sqrt{yz}\leq y+z.$$Thus, $x,y,z$ are the length of sides of some triangle. However, the area of that triangle determined by Heron’s are $$\sqrt{\frac{2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4}{16}}=\frac{\sqrt{2(ab+bc+ca)-(a^2+b^2+c^2)}}{4}$$which must be greater than $0$, as desired.

oVlad wrote: The positive numbers $a, b, c$ are such that $a^2<16bc, b^2<16ca$ and $c^2<16ab$. Prove that \[a^2+b^2+c^2<2(ab+bc+ca).\]

Attachments:

Quidditch wrote: Lol. Let $a=x^2,b=y^2,c=z^2$. Then $a^2<16bc$ implies that $$x^4<16y^2z^2\iff x<2\sqrt{yz}\leq y+z.$$Thus, $x,y,z$ are the length of sides of some triangle. However, the area of that triangle determined by Heron’s are $$\sqrt{\frac{2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4}{16}}=\frac{\sqrt{2(ab+bc+ca)-(a^2+b^2+c^2)}}{4}$$which must be greater than $0$, as desired. wow nice and cool solution