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For what values of $k>0$ does $$a^2+b^2+c^2+d^2+4(\sqrt3 -1)(abcd)^k\geq\sqrt{12(abc+abd+acd+bcd)}$$ hold for all $a,b,c,d\geq0$ satisfying $a+b+c+d=4$?
On the one hand, I found a lower bound for the LHS by using an equivalent form of Turkevich's inequality : $3(a^2+b^2+c^2+d^2)\geq4(4-abcd)$.
On the other hand, for the RHS one can find an upper bound by using the famous ISL 1997 $64+44abcd\geq27(abc+abd+acd+bcd)$.
Yet, I am stuck, I cannot make any progress towards finding the range of $k$.

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    $\begingroup$ I have a proof for $k=1$. For $k=\frac{200}{199}$ your inequality is wrong. $\endgroup$ Commented Dec 3, 2019 at 18:55
  • $\begingroup$ @MichaelRozenberg Nice ! I also suspected that $k_{max}=1$, but I couldn't prove that. $\endgroup$
    – Math Guy
    Commented Dec 3, 2019 at 19:13
  • $\begingroup$ @ Math Guy, are you sure that your "suspicion" is right? What tells us that $k_{max}=1$ ? $\endgroup$
    – user120123
    Commented Dec 3, 2019 at 19:20
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    $\begingroup$ Thank you for posting! I'm very glad for seeing 3 of my proposals posted here. I suppose I'm not allowed to participate at this bounty haunting. $\endgroup$
    – user120123
    Commented Dec 5, 2019 at 19:49

1 Answer 1

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Let $k>1$ and $b=c=a$.

Thus, $d=4-3a$, $0\leq a\leq\frac{4}{3}$ and we need $$3a^2+(4-3a)^2+4(\sqrt3-1)(a^3(4-3a))^k\geq2\sqrt{3(a^3+3a^2(4-3a))}$$ or $$3a^2-6a+4-a\sqrt{3(3-2a)}+(\sqrt3-1)(a^3(4-3a))^k\geq0$$ or $$\frac{(4-3a)(4-9a+6a^2-3a^3)}{3a^2-6a+4+a\sqrt{3(3-2a)}}+(\sqrt3-1)(a^3(4-3a))^k\geq0$$ or $$\frac{4-9a+6a^2-3a^3}{3a^2-6a+4+a\sqrt{3(3-2a)}}+(\sqrt3-1)a^{3k}(4-3a)^{k-1}\geq0,$$ which is wrong for $a=\frac{4}{3},$ which says that $k\leq1$.

We'll prove that $k=1$ is valid.

Indeed, let $a=\min\{a,b,c,d\}$ and $$f(a,b,c,d)=a^2+b^2+c^2+d^2+4(\sqrt3-1)abcd-2\sqrt{3(abc+abd+acd+bcd)}.$$

Thus, $0\leq a\leq1$ and $$f(a,b,c,d)-f\left(a,\frac{b+c+d}{3},\frac{b+c+d}{3},\frac{b+c+d}{3}\right)=$$ $$=b^2+c^2+d^2-\frac{(b+c+d)^2}{3}-4(\sqrt3-1)a\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)+$$ $$+2\sqrt3\left(\sqrt{\frac{a(b+c+d)^2}{3}+\frac{(b+c+d)^3}{27}}-\sqrt{abc+abd+acd+bcd}\right)=$$ $$=\frac{2}{3}\sum_{cyc}(b^2-bc)-4(\sqrt3-1)a\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)+$$ $$+\frac{\frac{2a}{\sqrt3}\sum\limits_{cyc}(b^2-bc)+2\sqrt3\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)}{\sqrt{\frac{a(b+c+d)^2}{3}+\frac{(b+c+d)^3}{27}}+\sqrt{abc+abd+acd+bcd}}\geq$$ $$\geq\frac{2}{3}\sum_{cyc}(b^2-bc)-4(\sqrt3-1)a\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)+$$ $$+\frac{\frac{2a}{\sqrt3}\sum\limits_{cyc}(b^2-bc)+2\sqrt3\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right)}{4}=$$ $$=\left(\frac{2}{3}+\frac{a}{2\sqrt3}\right)\sum_{cyc}(b^2-bc)+\left(\frac{\sqrt3}{2}-4(\sqrt3-1)a\right)\left(\left(\frac{b+c+d}{3}\right)^3-bcd\right).$$ Now, if $\frac{\sqrt3}{2}-4(\sqrt3-1)a\geq0$ so $f(a,b,c,d)-f\left(a,\frac{b+c+d}{3},\frac{b+c+d}{3},\frac{b+c+d}{3}\right)\geq0.$

But for $\frac{\sqrt3}{2}-4(\sqrt3-1)a\leq0$ since by Schur $$\left(\frac{b+c+d}{3}\right)^3-bcd\leq\frac{4}{27}\sum_{cyc}(b^3-bcd),$$ we obtain: $$f(a,b,c,d)-f\left(a,\frac{b+c+d}{3},\frac{b+c+d}{3},\frac{b+c+d}{3}\right)\geq$$ $$\geq\left(\frac{2}{3}+\frac{a}{2\sqrt3}\right)\sum_{cyc}(b^2-bc)+\left(\frac{\sqrt3}{2}-4(\sqrt3-1)a\right)\cdot\frac{4}{27}\sum_{cyc}(b^3-bcd)=$$ $$=\left(\frac{2}{3}+\frac{a}{2\sqrt3}+\left(\frac{\sqrt3}{2}-4(\sqrt3-1)a\right)\cdot\frac{4}{27}(4-a)\right)\sum_{cyc}(b^2-bc)\geq0$$ for any $0\leq a\leq1.$

Id est, $$f(a,b,c,d)\geq f\left(a,\frac{b+c+d}{3},\frac{b+c+d}{3},\frac{b+c+d}{3}\right),$$ which says that it's enough to prove our inequality for equality case of three variables.

Let $b=c=a$.

Thus, $d=4-3a$, where $0\leq a\leq \frac{4}{3}$ and we need to prove that: $$3a^2+(4-3a)^2+4(\sqrt3-1)a^3(4-3a)\geq2\sqrt{3(a^3+3a^2(4-3a))},$$ which after squaring of the both sides gives: $$(a-1)^2(4-3a)(4-a+2(4\sqrt3-5)a^3-4(2-\sqrt3)a^4-6(2-\sqrt3)a^5)\geq0,$$ which is true for $0\leq a\leq \frac{4}{3}$.

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    $\begingroup$ For $k>1$, I would have simplified that factor first and then I would have taken the limit. Anyway, the proof is good, congratulations ! $\endgroup$
    – Math Guy
    Commented Dec 6, 2019 at 22:16
  • $\begingroup$ what's "Id est,"? $\endgroup$ Commented Dec 12, 2019 at 0:40
  • $\begingroup$ @mathworker21 Id est=i.e. $\endgroup$ Commented Dec 12, 2019 at 4:54

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