Generalized of this problem.

MRF2017 wrote: For $a,b,c,d>0$ prove that $$\frac {a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4} \leq \sqrt[4]{a.\frac{a+b}{2}.\frac{a+b+c}{3}.\frac{a+b+c+d}{4}}$$

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MRF2017 wrote: For $a,b,c,d>0$ prove that $$\frac {a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4} \leq \sqrt[4]{a.\frac{a+b}{2}.\frac{a+b+c}{3}.\frac{a+b+c+d}{4}}$$ $a \cdot \frac{{a + b}}{2} \cdot \frac{{a + b + c}}{3} \cdot \frac{{a + b + c + d}}{4} = $ $\frac{1}{4}\left( {a + a + a + a} \right) \cdot \frac{1}{4}\left( {a + \frac{{2a + b}}{3} + \frac{{a + 2b}}{3} + b} \right) \cdot \frac{1}{4}\left( {a + \frac{{a + 2b}}{3} + \frac{{2b + c}}{3} + c} \right) \cdot \frac{1}{4}\left( {a + b + c + d} \right) \ge $ $\frac{1}{{4^4 }}\left( {a + a + a + a} \right) \cdot \left( {a + \sqrt[3]{{a^2 b}} + \sqrt[3]{{ab^2 }} + b} \right) \cdot \frac{1}{4}\left( {a + \sqrt[3]{{ab^2 }} + \sqrt[3]{{b^2 c}} + c} \right) \cdot \frac{1}{4}\left( {a + b + c + d} \right) \ge $ $\frac{1}{{4^4 }}\left( {\sqrt[4]{{aaaa}} + \sqrt[4]{{a \cdot \sqrt[3]{{a^2 b}} \cdot \sqrt[3]{{ab^2 }} \cdot b}} + \sqrt[4]{{a \cdot \sqrt[3]{{ab^2 }} \cdot \sqrt[3]{{b^2 c}} \cdot c}} + \sqrt[4]{{abcd}}} \right)^4 = $ $\left( {\frac{{a + \sqrt {ab} + \sqrt[3]{{abc}} + \sqrt[4]{{abcd}}}}{4}} \right)^4 $

We can use Holder like this \[\dfrac{a+a+a+a}{4}.\dfrac{a+a+b+b}{4}.\dfrac{a+b+\sqrt[3]{abc}+c}{4}.\dfrac{a+b+c+d}{4}\geq \left(\dfrac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\right)^4\]Note that $4(a+b+c)\geq 3(a+b+c+\sqrt[3]{abc})$ and the proof is completed. I still can't solve the general inequality in case $k=5$, it's over 1 year from the first time I know it

The general case proof can be found here http://artofproblemsolving.com/community/c6h305822 and here http://artofproblemsolving.com/community/c6h6474

hoanglong2k wrote: I still can't solve the general inequality in case $k=5$, it's over 1 year from the first time I know it Did you try using sqing's method?

let RHS=4S,\[\begin{array}{l} \frac{a}{{S}} \le 1 + \frac{{2a}}{{a + b}} + \frac{{3a}}{{a + b + c}} + \frac{{4a}}{{a + b + c + d}}\\\\ \frac{{\sqrt {ab} }}{{S}} \le 1 + \frac{{2a}}{{a + b}} + \frac{{3b}}{{a + b + c}} + \frac{{4b}}{{a + b + c + d}}\\\\ \frac{{{}^3\sqrt {abc} }}{{S}} \le 1 + \frac{{2b}}{{a + b}} + \frac{{3{}^3\sqrt {abc} }}{{a + b + c}} + \frac{{4c}}{{a + b + c + d}}\\\\ \frac{{{}^4\sqrt {abcd} }}{{S}} \le 1 + \frac{{2b}}{{a + b}} + \frac{{3c}}{{a + b + c}} + \frac{{4d}}{{a + b + c + d}} \end{array}\]add above all, we complete our proof by using AM-GM inequality

not mine , but Mr.chenji , this new problem is much more difficult than the Bulgaria one , i think.attachment

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By Hölder, we have $(a+\sqrt{ab}+{}^3\sqrt{abc}+{}^4\sqrt{abcd})^4\le(a+a+a+a)(a+a+b+b)(a+b+{}^3\sqrt{abc}+c)(a+b+c+d)$. And the result follows.

Andyqian7 wrote: By Hölder, we have $(a+\sqrt{ab}+{}^3\sqrt{abc}+{}^4\sqrt{abcd})^4\le(a+a+a+a)(a+a+b+b)(a+b+{}^3\sqrt{abc}+c)(a+b+c+d)$. And the result follows. Solution is almost same as #8.

Pluto04 wrote: Andyqian7 wrote: By Hölder, we have $(a+\sqrt{ab}+{}^3\sqrt{abc}+{}^4\sqrt{abcd})^4\le(a+a+a+a)(a+a+b+b)(a+b+{}^3\sqrt{abc}+c)(a+b+c+d)$. And the result follows. Solution is almost same as #8. Sorry I didn't note this.

Really cool problem! This inequality is immediately trivial by this.(Kedlayas result) LHS$\leq\sqrt[4]{a.\frac{a+b}{2}.\frac{a+b+c}{3}.\frac{a+b+c+d}{4}}$ (by Kedlayas result) And this immediately finishes our problem!