# 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.

• Just an aside. If $a,b,c,$ can be the sides of a triangle, let $s=(a+b+c)/2,$ and use Heron's formula to show the desired inequality reduces to $a^2-2ab\sin\theta+b^2\ge0,$ which is trivial. I wonder if there's an easy way to handle the non-triangle case. – saulspatz Apr 27 '18 at 1:56
• @saulspatz: Neat! If not the sides of a triangle, then the right side will be negative or 0 I believe. So perhaps you would like to add an answer with your nice proof? – Aryabhata Apr 27 '18 at 2:02
• @Aryabhata Done. – saulspatz Apr 27 '18 at 2:37
• @saulspatz: Thanks! Upvoted. – Aryabhata Apr 28 '18 at 1:33
• @Abe Are you sure that discussion of problems online from past BMO is legal? I have seen a post by sir Geoff Smith(can't remember the source now), he said not to discuss BMO questions online,motive was to force later BMO participants to do the problem by themselves. – tarit goswami Oct 10 '18 at 2:47

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 $c^2(a^2+b^2)\neq0$.

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

• This may be self evident but would you be able to explain how AM-GM is used in the first inequality? – Abe Apr 27 '18 at 6:52
• @Abe I added something. See now. – Michael Rozenberg Apr 27 '18 at 8:59

Suppose that $a,b,c$ can form the sides of a triangle. Let $s=\frac{a+b+c}{2}$ be the semiperimeter. The inequality becomes $$(a^2+b^2)^2\ge2s\cdot 2(s-a)\cdot 2(s-b)\cdot 2(s-c)$$ or by Heron's formula, $$a^2+b^2\ge 4A$$ where $A$ is the area of the triangle. If $\theta$ is the angle between sides $a$ and $b$, this reduces to $$a^2-2ab\sin\theta+b^2\ge0,$$ and we have$$a^2-2ab\sin\theta+b^2\ge a^2-2ab+b^2=(a-b)^2\ge0$$

In the case where $a,b,c$ do not form a triangle, exactly one of the factors on the right-hand is negative or one of the factors is $0$, so the inequality is trivial.

Let's simplify the inequality: $$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b) \Rightarrow \\ a^4+2a^2b^2+b^4\ge ((a+b)^2-c^2)(c^2-(a-b)^2) \Rightarrow \\ a^4+2a^2b^2+b^4\ge 2c^2(a^2+b^2)-(a^2-b^2)^2-c^4 \Rightarrow \\ c^4-2(a^2+b^2)c^2+2(a^4+b^4)\ge 0 \qquad (1)$$ It is a bi-quadratic inequality and its discriminant is: $$D=(a^2+b^2)^2-2(a^4+b^4)=2a^2b^2-(a^4+b^4)\le 0,$$ the iquality occurs for $a=b$. Note that the inequlity $(1)$ is true for $D<0$. We will check $D=0$. Then the inequality $(1)$ and its solution will be: $$\begin{cases} c^2=a^2+b^2=2a^2 \\ (2a^2)^4-4a^4+4a^4\ge 0\end{cases} \Rightarrow16a^8\ge 0.$$