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Without losing generality we assume that ab≥0. From the 1st condition we have a+b=−c replace this into the 2nd condition
a2+b2+(a+b)2=1
By C-S a2+b2≥2(a+b)2, so 1≥23(a+b)2⇒c2≤32. Now we rewrite the 2nd condition
2(a2+b2)+2ab=1
We have this inequality for all reals a2+b2≥2ab, therefore 6ab≤1⇒a2b2≤361∴a2b2c2≤541
The equality holds when (a,b,c)=(61,61,6−2);(6−1,6−1,62)
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Without losing generality we assume that ab≥0. From the 1st condition we have a+b=−c replace this into the 2nd condition a2+b2+(a+b)2=1 By C-S a2+b2≥2(a+b)2, so 1≥23(a+b)2⇒c2≤32. Now we rewrite the 2nd condition 2(a2+b2)+2ab=1 We have this inequality for all reals a2+b2≥2ab, therefore 6ab≤1⇒a2b2≤361 ∴a2b2c2≤541 The equality holds when (a,b,c)=(61,61,6−2);(6−1,6−1,62)
Let α=abc.
Now, as 0=(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⟹0=1+2(ab+bc+ca)⟹ab+bc+ca=−21
Therefore, a,b,c are the (real) roots of the following function:
f(x)=x3−21x−α
For the existence of three real roots, we must have: f(x1)f(x2)≤0 where x1 and x2 are the points of extrema of the function f(x).
Note that f′(x1)=f′(x2)=0.
Since f′(x) is a quadratic polynamial, it can be solved.
Solving, plugging it into the above inequality, we get: ∣α∣≤361⟹α2≤541
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.