# Not true!

Let $a,b,c$ be real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1.$Prove that $a^2b^2c^2\leq \frac{1}{54}.$When does this equality hold true?

Note by Ayush G Rai
3 years, 11 months ago

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Without losing generality we assume that $ab\geq 0$. From the 1st condition we have $a+b=-c$ replace this into the 2nd condition $a^2+b^2+(a+b)^2=1$ By C-S $a^2+b^2\geq\frac{(a+b)^2}{2}$, so $1\geq\frac{3}{2}(a+b)^2 \Rightarrow c^2\leq\frac{2}{3}$. Now we rewrite the 2nd condition $2(a^2+b^2)+2ab=1$ We have this inequality for all reals $a^2+b^2\geq 2ab$, therefore $6ab\leq 1\Rightarrow a^2b^2\leq\frac{1}{36}$ $\therefore a^2b^2c^2\leq\frac{1}{54}$ The equality holds when $(a,b,c)=\bigg(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\bigg);\bigg(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}\bigg)$

- 3 years, 11 months ago

Let $\alpha=abc$.

Now, as $0=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\\implies 0=1+2(ab+bc+ca)\\\implies ab+bc+ca=-\frac{1}{2}$

Therefore, $a, b, c$ are the (real) roots of the following function:

$f(x)=x^3-\frac{1}{2}x-\alpha$

For the existence of three real roots, we must have: $f(x_1)f(x_2) \leq 0$ where $x_1$ and $x_2$ are the points of extrema of the function $f(x)$.

Note that $f'(x_1)=f'(x_2)=0$.

Since $f'(x)$ is a quadratic polynamial, it can be solved.

Solving, plugging it into the above inequality, we get: $|\alpha| \leq \frac{1}{3\sqrt6} \\\implies \alpha^2 \leq \frac{1}{54}$

With equality holding if and only if two of the real numbers $a, b, c$ are equal.

- 3 years, 11 months ago

Note: It's slightly more straightforward to just take the cubic discriminant condition for 3 real roots.

Staff - 3 years, 11 months ago

The problem was I didn't remeber the cubic discriminant formula then. So, I used calculus in the solution.

- 3 years, 11 months ago