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?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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 \((a,b,c)=\bigg(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\bigg);\bigg(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}\bigg)\) – Gurīdo Cuong · 1 year, 2 months ago

Log in to reply

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. – Deeparaj Bhat · 1 year, 2 months ago

Log in to reply

cubic discriminant condition for 3 real roots. – Calvin Lin Staff · 1 year, 2 months ago

Note: It's slightly more straightforward to just take theLog in to reply

– Deeparaj Bhat · 1 year, 2 months ago

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

Log in to reply

– Deeparaj Bhat · 1 year, 2 months ago

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

Since \(a, b, c\) are real, \(a^2b^2c^2 \geq 0\). – Deeparaj Bhat · 1 year, 2 months ago

Log in to reply

I got a question... if you post a problem as a note, why to give yourself the solution? – Mateo Matijasevick · 1 year, 2 months ago

Log in to reply

– Ayush Rai · 1 year, 2 months ago

because it is a proving problemLog in to reply

– Abhishek Alva · 1 year, 2 months ago

ya you are rightLog in to reply

– Ayush Rai · 1 year, 2 months ago

i know you did not undertand anything.Just interrupting for nothing.Log in to reply

@Deeparaj Bhat – Ayush Rai · 1 year, 2 months ago

okLog in to reply