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Let $a,b,c,d \in \Bbb R$, $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$. Then $$36 \leq 4(a^3+b^3+c^3+d^3) - (a^4+b^4+c^4+d^4)\leq4 8.$$

I have found only two bounds: $216 \geq a^3+b^3+c^3+d^3$ and $144 \geq a^4+b^4+c^4+d^4$.

How to prove this inequality?

nonuser
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User8976
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    How did you find the upper bounds? – Mr Pie Feb 27 '18 at 11:27
  • I think we can use the EV method . https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf . Particulary the corollary 1.5 .Note that it works just for non-negative numbers – max8128 Feb 27 '18 at 11:39
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    What is the source of this problem? I hate to sound paranoid, but some users have tried to cheat in contests, and questions of this type just might be from a contest somewhere. – Jyrki Lahtonen Feb 27 '18 at 11:43
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    @max8128 The method of Lagrange multipliers implies easily that all the variables are in the interval $[0,3]$. The extrema being attained at permutations of $(a,b,c,d)=(0,2,2,2)$ and $(3,1,1,1)$ respectively. It won't surprise me one bit, if non-negativity can be shown by even more elementary means. – Jyrki Lahtonen Feb 27 '18 at 11:47

1 Answers1

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For upper bound, we have

$$ x^3(4-x) \leq 4x^2\;\; \forall x$$

and we are done.

For lower bound:

Let $a+b+c =x$, then $d=6-x$ and by Cauchy we have $a^2+b^2+c^2\geq x^2/3$

so $$12 \geq {x^2\over 3} +(6-x)^2$$ so $x\in (3,6) $ and so $0\leq d\leq 3$. But this holds also for $a,b,c$.

Idea for finish:

Let $f(x) = 4x^3-x^4$, for $x\in[0,3]$ we have $4x^3-x^4\geq x^3$, so we have by Cauchy $$ f(a)+f(b)+f(c)\geq a^3+b^3+c^3\geq {(a^2+ b^2+c^2)^2\over a+b+c} = {(12-d^2)^2\over 6-d}$$

So we have to check if following holds:

$$ f(d)+ {(12-d^2)^2\over 6-d}\geq 36$$

nonuser
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