6
$\begingroup$

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 $k$.

$\endgroup$
4
  • 1
    $\begingroup$ I have a proof for $k=1$. For $k=\frac{200}{199}$ your inequality is wrong. $\endgroup$ – Michael Rozenberg Dec 3 '19 at 18:55
  • $\begingroup$ @MichaelRozenberg Nice ! I also suspected that $k_{max}=1$, but I couldn't prove that. $\endgroup$ – Math Guy Dec 3 '19 at 19:13
  • $\begingroup$ @ Math Guy, are you sure that your "suspicion" is right? What tells us that $k_{max}=1$ ? $\endgroup$ – user120123 Dec 3 '19 at 19:20
  • 1
    $\begingroup$ 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 Dec 5 '19 at 19:49
5
+500
$\begingroup$

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 $0\leq a\leq \frac{4}{3}$.

$\endgroup$
3
  • 2
    $\begingroup$ 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 Dec 6 '19 at 22:16
  • $\begingroup$ what's "Id est,"? $\endgroup$ – mathworker21 Dec 12 '19 at 0:40
  • $\begingroup$ @mathworker21 Id est=i.e. $\endgroup$ – Michael Rozenberg Dec 12 '19 at 4:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.