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 \(a,b,c\) are positive reals,the equality holds when \(a=b=c\) and at that time,\(k=\frac{\sqrt{3}}{\sqrt{2}}\)

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

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

Let \(f(t)=\frac{(x+y+z-3t)^2}{(x-t)(y-t)(z-t)}\) is decreasing and \(g(t)=\frac{5\sqrt{2}(x+y+z-3t)\sqrt{(x-t)^2+(y-t)^2+(z-t)^2}}{4}-(x+y+z-3t)^2\) is increasing in \([0;4]\). It's true because if we let \(m=x-t;n=y-t,p=z-t\),\(m,n,p > 0\) and \(A=\sqrt{m^2+n^2+p^2};B=m+n+p\),we get: \(m^2n^2p^2f'(t) \geq 0\) and \(g'(t) \leq 0 \). So that we get: \[\frac{a}{\sqrt{a+b}}+\sqrt{b}\leq \frac{5}{4}\sqrt{a+b}\] This it's right because \(RHS-LHS \geq 0\). The problem is solved. Another way is more interested:By C-S Inequality \[LHS^2\leq \sum a(5a+b+9c)\sum \frac{a}{(a+b)(5a+b+9c)}\] We need to prove: \[(a+b+c)\sum \frac{a}{(a+b)(5a+b+9c)}\leq \frac{5}{16}\] This inequality right because: \[(RHS-LHS)(16\prod (5a+b+9c))=\sum ab(a+b)(a+9b)(a-3b)^2+243\sum a^3b^2c+835\sum a^3bc^2+232\sum a^4bc+1230a^2b^2c^2\geq 0\]

The problem is solved.

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## Comments

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TopNewestCan you explain this clearer?

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

\[\sum_{\mathrm{cyc}} \frac{a^{2}+b^{2}}{a^{8}+b^{8}}\leq a^{3}+b^{3}+c^{3}\]

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@Gurīdo Cuong;@Daniel Liu;@Brian Charlesworth;@Pi Han Goh

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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.

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Also Vietnam TST

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