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# 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 Rai
7 months, 2 weeks 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 $$(a,b,c)=\bigg(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}\bigg);\bigg(\frac{-1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}}\bigg)$$ · 7 months, 1 week 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. · 7 months, 2 weeks ago

Note: It's slightly more straightforward to just take the cubic discriminant condition for 3 real roots. Staff · 7 months, 2 weeks ago

The problem was I didn't remeber the cubic discriminant formula then. So, I used calculus in the solution. · 7 months, 2 weeks ago

Comment deleted 7 months ago

Btw, equality holds only if two of the three numbers $$a, b, c$$ are equal. · 7 months, 2 weeks ago

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, $$a^2b^2c^2 \geq 0$$. · 7 months, 2 weeks ago

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? · 7 months, 2 weeks ago

because it is a proving problem · 7 months, 1 week ago

ya you are right · 7 months, 2 weeks ago

i know you did not undertand anything.Just interrupting for nothing. · 7 months, 2 weeks ago