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 \((a,b,c)=\bigg(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\bigg);\bigg(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}\bigg)\)

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.

Right, actually, if the three are non-negative, then the three are equal to 0. But this can be proved by cases, for example, WLOG let's assume \(a\) and \(b\) are positive and \(c\) is negative, then \(-c\) is positive, and so on.

I got a question... if you post a problem as a note, why to give yourself the solution?

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TopNewestWithout 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 \((a,b,c)=\bigg(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\bigg);\bigg(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}\bigg)\)

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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.

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

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

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Comment deleted Jun 07, 2016

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Btw, equality holds only if two of the three numbers \(a, b, c\) are equal.

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You can't use am-gm inequality as \(a, b\) and \(c\) aren't non negative (if they are, then the inequality to be proved holds trivially).

Since \(a, b, c\) are real, \(a^2b^2c^2 \geq 0\).

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Right, actually, if the three are non-negative, then the three are equal to 0. But this can be proved by cases, for example, WLOG let's assume \(a\) and \(b\) are positive and \(c\) is negative, then \(-c\) is positive, and so on.

I got a question... if you post a problem as a note, why to give yourself the solution?

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ya you are right

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ok @Deeparaj Bhat

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