Daniel Liu's Hard Inequality

Hello every body.Sorry for too late.But this PROBLEM is very hard. In 1989,Jack Garfunkel recommended a inequality like this problem to Crux magazine in Canada.It's a hard problem and didn't have any solution at that time.

If we given a,b,ca,b,c are positive reals,the equality holds when a=b=ca=b=c and at that time,k=32k=\frac{\sqrt{3}}{\sqrt{2}}

But k=54k=\frac{5}{4} must be better but we must let a,b,ca,b,c are non-negative and the equality holds when a=b=3,c=0a=b=3,c=0 or it's permutation.

At present,there have many way to prove Jack Garfunkel's inequality. By applying full incremental variable,we have: a+b+c+(ab)(bc)(ac)(a+b+c)abc524x2+y2+z2a+b+c+\frac{(a-b)(b-c)(a-c)(a+b+c)}{abc}\leq \frac{5\sqrt{2}}{4}\sqrt{x^2+y^2+z^2}

Let f(t)=(x+y+z3t)2(xt)(yt)(zt)f(t)=\frac{(x+y+z-3t)^2}{(x-t)(y-t)(z-t)} is decreasing and g(t)=52(x+y+z3t)(xt)2+(yt)2+(zt)24(x+y+z3t)2g(t)=\frac{5\sqrt{2}(x+y+z-3t)\sqrt{(x-t)^2+(y-t)^2+(z-t)^2}}{4}-(x+y+z-3t)^2 is increasing in [0;4][0;4]. It's true because if we let m=xt;n=yt,p=ztm=x-t;n=y-t,p=z-t,m,n,p>0m,n,p > 0 and A=m2+n2+p2;B=m+n+pA=\sqrt{m^2+n^2+p^2};B=m+n+p,we get: m2n2p2f(t)0m^2n^2p^2f'(t) \geq 0 and g(t)0g'(t) \leq 0 . So that we get: aa+b+b54a+b\frac{a}{\sqrt{a+b}}+\sqrt{b}\leq \frac{5}{4}\sqrt{a+b} This it's right because RHSLHS0RHS-LHS \geq 0. The problem is solved. Another way is more interested:By C-S Inequality LHS2a(5a+b+9c)a(a+b)(5a+b+9c)LHS^2\leq \sum a(5a+b+9c)\sum \frac{a}{(a+b)(5a+b+9c)} We need to prove: (a+b+c)a(a+b)(5a+b+9c)516(a+b+c)\sum \frac{a}{(a+b)(5a+b+9c)}\leq \frac{5}{16} This inequality right because: (RHSLHS)(16(5a+b+9c))=ab(a+b)(a+9b)(a3b)2+243a3b2c+835a3bc2+232a4bc+1230a2b2c20(RHS-LHS)(16\prod (5a+b+9c))=\sum ab(a+b)(a+9b)(a-3b)^2+243\sum a^3b^2c+835\sum a^3bc^2+232\sum a^4bc+1230a^2b^2c^2\geq 0

The problem is solved.

Note by Laurent Michael
3 years, 7 months ago

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Can you explain this clearer?

  1. What are you trying to describe?
  2. What is the process to solve such a problem
  3. How does someone come up with the various inequalities?

Calvin Lin Staff - 3 years, 6 months ago

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Prove this Consider a,b,c>0,abc=1a,b,c>0, abc=1. Prove that

cyca2+b2a8+b8a3+b3+c3\sum_{\mathrm{cyc}} \frac{a^{2}+b^{2}}{a^{8}+b^{8}}\leq a^{3}+b^{3}+c^{3}

Shivam Jadhav - 3 years, 7 months ago

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Ah yes, this was the SOS method you mentioned on the solutions page right? I believe it is from a China TST, as well.

Ameya Daigavane - 3 years, 7 months ago

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Also Vietnam TST

Laurent Michael - 3 years, 7 months ago

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