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Find the largest value of $\alpha$ such that

$$\left(a_1^2+a_2^2+a_3^2+a_4^2\right)^3\ge\alpha\left(a_1^3+a_2^3+a_3^3+a_4^3\right)^2$$

where $a_1,a_2,a_3$ and $a_4$ are real numbers and $a_1+a_2+a_3+a_4=0$.

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  • $\begingroup$ You seem to be seeking the maximum $\alpha$ for which this inequality holds true for all real $a_i$. You also need to mention what you have done so far on the problem, so that appropriate help can be given. If I just say $\alpha_{max} = 3$, that wouldn't be very helpful, would it? $\endgroup$ – Macavity Jan 6 '17 at 5:16
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Hints:. If all $a_i$ are zero, then clearly any $\alpha$ works. Else, we may use homogeneity to have a suitable additional constraint - in particular, we may equivalently seek the maximum of $\sum a_i^3$ under the constraints $\sum a_i = 0$ (given) and $\sum a_i^2=1$ (new from homogeneity).

With these constraints, we may note that $a_i$ are roots of a quartic $p(x) = x^4-\frac12x^2-p x + q$, where $p, q \in \mathbb R$ and we seek the maximum of $\sum a_i^3 = 3p$. As all roots are real, $p'(x) = 4x^3-x -p$ must have non-negative discriminant, so $\Delta = 16-27\times4^2 p^2 \geqslant 0 \implies p \leqslant \frac1{3\sqrt3}$.

Note equality is possible, so $\sum a_i^3 = 3p$ has a maximum of $\frac1{\sqrt3}$, leading us to: $$\sum a_i = 0, \sum a_i^2=1 \quad \implies \sum a_i^3 \leqslant \frac1{\sqrt3}$$ $$\therefore \sum a_i = 0 \quad \implies \left(\sum a_i^2 \right)^3 \geqslant 3 \left(\sum a_i^3 \right)^2$$ or $\alpha_{max} = 3$.

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  • $\begingroup$ I am just curious: Can you explicitly determine values $a_1, ..., a_4$ for which the quotient is equal to $3$? $\endgroup$ – Martin R Jan 6 '17 at 6:43
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    $\begingroup$ @MartinR Try $-1, -1, -1, 3$ for instance... Discriminant of $p'(x)$ becoming zero needs multiplicity of roots, so the optimal forms are often of this form. $\endgroup$ – Macavity Jan 6 '17 at 6:53
  • $\begingroup$ Does a similar method work with math.stackexchange.com/q/2081528/42969? I am fairly sure that the result is correct and equality attained for (multiples and permutations of) $(3, 3, -2, -2, -2)$ but I was not able to prove it yet. $\endgroup$ – Martin R Jan 6 '17 at 6:59
  • $\begingroup$ @MartinR will check, though discriminant of that order doesn't look easy to check or manipulate... $\endgroup$ – Macavity Jan 6 '17 at 7:10
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There might not exist a such largest $\alpha$.

In fact, suppose $a_1+a_2=0$. Then $a_1^3+a_2^3=0$, and since $a_1+a_2+a_3+a_4=0$, we get $a_3+a_4=0$, which gives $a_3^3+a_4^3=0$. That is, the RHS of the inequality is vanishing, and then the inequality holds regardless of $\alpha$.

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  • $\begingroup$ $\alpha$ is supposed to be a constant satisfying the inequality, if I'm not mistaken. $\endgroup$ – Blencer Jan 5 '17 at 23:46
  • $\begingroup$ @ Guru: I had missed that when I was half-way of my typing. Now what is left that is a fact. $\endgroup$ – MathChat Jan 6 '17 at 0:01

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