Not true!

Let a,b,ca,b,c be real numbers such that a+b+c=0a+b+c=0 and a2+b2+c2=1.a^2+b^2+c^2=1.Prove that a2b2c2154.a^2b^2c^2\leq \frac{1}{54}.When does this equality hold true?

Note by Ayush G Rai
5 years ago

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Without losing generality we assume that ab0ab\geq 0 . From the 1st condition we have a+b=ca+b=-c replace this into the 2nd condition a2+b2+(a+b)2=1a^2+b^2+(a+b)^2=1 By C-S a2+b2(a+b)22a^2+b^2\geq\frac{(a+b)^2}{2}, so 132(a+b)2c2231\geq\frac{3}{2}(a+b)^2 \Rightarrow c^2\leq\frac{2}{3}. Now we rewrite the 2nd condition 2(a2+b2)+2ab=12(a^2+b^2)+2ab=1 We have this inequality for all reals a2+b22aba^2+b^2\geq 2ab, therefore 6ab1a2b21366ab\leq 1\Rightarrow a^2b^2\leq\frac{1}{36} a2b2c2154\therefore a^2b^2c^2\leq\frac{1}{54} The equality holds when (a,b,c)=(16,16,26);(16,16,26)(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)

P C - 5 years ago

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Let α=abc\alpha=abc.

Now, as 0=(a+b+c)2=a2+b2+c2+2(ab+bc+ca)    0=1+2(ab+bc+ca)    ab+bc+ca=120=(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,ca, b, c are the (real) roots of the following function:


For the existence of three real roots, we must have: f(x1)f(x2)0f(x_1)f(x_2) \leq 0 where x1x_1 and x2x_2 are the points of extrema of the function f(x)f(x).

Note that f(x1)=f(x2)=0f'(x_1)=f'(x_2)=0.

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

Solving, plugging it into the above inequality, we get: α136    α2154|\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,ca, 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.

Calvin Lin Staff - 5 years ago

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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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