British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states For positive real numbers $a,b,c$ prove that $(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$ After some algebraic wrangling we can get to the point where: $(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥ 2c^2(a^2 + b^2)$ At this point if we take the $LHS - RHS$ we can write the expression as the sum of squares proving the inequality. I was wondering, is it possible to divide both sides by $c^2(a^2 + b^2)$ and show somehow that $((a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4)/(c^2(a^2 + b^2)) ≥ 2$ I tried but was not able to.

We need to prove that $$(a^2+b^2)^2\geq\sum_{cyc}(2a^2b^2-a^4)$$ or $$c^4-2(a^2+b^2)c^2+2(a^4+b^4)\geq0$$ or $$(c^2-a^2-b^2)^2+(a^2-b^2)^2\geq0.$$

Yes, you can prove this inequality by the dividing.

Indeed, if $c^2(a^2+b^2)=0$ then the inequality is obvious.

Let $c^2(a^2+b^2)\
eq0$.

Thus, by AM-GM and Cauchy-Schwarz $$\frac{c^4+2(a^4+b^4)}{c^2(a^2+b^2)}=\frac{c^2}{a^2+b^2}+\frac{2(a^4+b^4)}{c^2(a^2+b^2}\geq$$ $$\geq2\sqrt{\frac{c^2}{a^2+b^2}\cdot\frac{2(a^4+b^4)}{c^2(a^2+b^2)}}=2\sqrt{\frac{2(a^4+b^4)}{(a^2+b^2)^2}}=$$ $$=2\sqrt{\frac{(1+1)(a^4+b^4)}{(a^2+b^2)^2}}\geq2\sqrt{\frac{(a^2+b^2)^2}{(a^2+b^2)^2}}=2.$$