There is a one-line solution to this problem. You'll have to find it on your own, though.

My problem here was the solution I submitted. I'll let someone else post the one-liner. Let $S$ be the desired quantity. The answer is \[ S \le \frac{8}{\sqrt[3]{7}} \quad \text{ achieved when } \quad (a,b,c,d) = (49, 1, 49, 1). \]We present three solutions for the upper bound. Set $a = w^2$, $b = x^2$, $c = y^2$, $d = z^2$ and moreover define $t = w+x+y+z$. First, by H\"older inequality we have \[ S^3 \le \left( \sum_{\text{cyc}} w \right) \left( \sum_{\text{cyc}} w \right) \left( \sum_{\text{cyc}} \frac{1}{x^2+7} \right). \]Also, we use will use the key estimate \begin{align*} 0 &\le \frac{(x-1)^2 (x-7)^2}{x^2+7} \\ &= x^2-16x+71 - \frac{448}{x^2+7} \\ \implies \frac{1}{x^2+7} &\le \frac{x^2-16x+71}{448} \\ \implies \sum_{\text{cyc}} \frac{1}{x^2+7} &\le \sum_{\text{cyc}} \frac{x^2-16x+71}{448} = \frac{6}{7} - \frac{t}{28} \end{align*}Consequently, the earlier estimate gives \[ S^3 \le \frac{t^2(24-t)}{28} = \frac{t \cdot t \cdot (48-2t)}{56} \le \frac{16^3}{56} = \frac{512}{7} \]as desired, with the last inequality by AM-GM.

v_Enhance wrote: I'll let someone else post the one-liner. If you insist. \[\sqrt[3]{\frac{7a}{64(b+7)}}\leq \frac{1}{3}\left(\frac{a+7}{64}+\frac{a}{a+7}+\frac{7}{b+7}\right).\quad \blacksquare\]Attributed to "Russian jury".

I think this problem is a very very very beautiful one and it's become one of my favourite inequality as well. I see all three above solution and I suddenly fell in love with them. I hope that I will see more solution. Thanks a lot to @v_Enhance who purpose this beautiful problem. My solution is very similar to @CantonMathGuy but my proof of the lemma that I found on the first time is quite different but I just realize later that it's very similar to the original one. (Indeed, my friend convinced me to show this proof on the forum. lol)

Solution by Gems98 We claim that the answer is $\sqrt[3]{\frac{512}{7}}$ achieved by $(a,b,c,d) = (49,1,49,1)$. Let $x,y,z,t$ be the permutation of $a,b,c,d$ in ascending order. From rearrangement inequality, we get $$S \leq \sqrt[3]{\frac{x}{t+7}} + \sqrt[3]{\frac{y}{z+7}} + \sqrt[3]{\frac{z}{y+7}} + \sqrt[3]{\frac{t}{x+7}}$$The key of the problem is the following claim. Claim: $\sqrt[3]{\dfrac{7a}{b+7}} + \sqrt[3]{\dfrac{7b}{a+7}} \leq \sqrt[3]{a+b+14}$ Proof: From Holder Inequality, we get \begin{align*} \sqrt[3]{7}\sqrt[3]{a}\sqrt[3]{\frac{1}{b+7}} + \sqrt[3]{b}\sqrt[3]{7}\sqrt[3]{\frac{1}{a+7}} &\leq \sqrt[3]{7+b}\sqrt[3]{a+7}\sqrt[3]{\frac{1}{b+7}+\frac{1}{a+7}} \\ &= \sqrt[3]{a+b+14} \end{align*}as claimed. Back to the main problem. Applying the claim, we get $S \leq \sqrt[3]{\frac{x+t+14}{7}} +\sqrt[3]{\frac{y+z+14}{7}}$. Hence from Power Mean inequality, we get $$\sqrt[3]{\frac{x+t+14}{7}} + \sqrt[3]{\frac{y+z+14}{7}} \leqslant \sqrt[3]{4\left(\frac{x+y+z+t+28}{7}\right)} = \sqrt[3]{\frac{512}{7}}$$as desired.

Solution by L. Hadassi and S. Regavim. By Holder and AM-GM we obtain: $$\sum_{cyc}\sqrt[3]{\frac{a}{b+7}}=\sum_{cyc}\sqrt[3]{\frac{a}{a+7}\cdot\frac{1}{b+7}\cdot(a+7)}\leq\sqrt[3]{\sum_{cyc}\frac{a}{a+7}\sum_{cyc}\frac{1}{b+7}\sum_{cyc}(a+7)}=$$$$=\sqrt[3]{128\left(4-7\sum_{cyc}\frac{1}{a+7}\right)\sum_{cyc}\frac{1}{a+7}}=\sqrt[3]{128\cdot7\left(\frac{4}{7}-\sum_{cyc}\frac{1}{a+7}\right)\sum_{cyc}\frac{1}{a+7}}\leq$$$$\leq\sqrt[3]{128\cdot7\cdot\left(\frac{2}{7}\right)^2}=\frac{8}{\sqrt[3]7}.$$

Pretty similar to idea used in 2008 Shortlist A5 Obvious that we shoud use Holder, After noticing that equality condition is $(a,b,c,d)=(1,49,1,49)$ $\left( \sum_{cyc}\sqrt[3]{\frac{a}{b+7}} \right)^3 \le \left( \sum_{cyc}(a+7) \right) \left( \sum_{cyc}\frac{a}{a+7} \right) \left( \sum_{cyc}\frac{1}{b+7} \right)$ Let $ \left( \sum_{cyc}\frac{a}{a+7} \right)=X$ and $\left( \sum_{cyc}\frac{1}{b+7} \right)=Y$ Note that $\left( \sum_{cyc}(a+7) \right)=128$ and $ X+7Y=\left( \sum_{cyc}\frac{a+7}{a+7} \right)=4 $ Therefore $2 \sqrt{7XY} \le 4 $ Which means that $XY \le \frac{4}{7}$ Finally $\left( \sum_{cyc}\sqrt[3]{\frac{a}{b+7}} \right)^3 \le \left( \sum_{cyc}(a+7) \right) \left( \sum_{cyc}\frac{a}{a+7} \right) \left( \sum_{cyc}\frac{1}{b+7} \right) \le 128 \times \frac{4}{7} = \frac{512}{7} $

Probably stronger, but same proof works I guess Prove that $$\sqrt[3]{\frac{a}{b+7}+\frac{c}{d+7}}+\sqrt[3]{\frac{b}{c+7}+\frac{d}{a+7}}\leq\sqrt[3]{\frac{128}{7}}$$where $a,b,c,d$ are non-negative real numbers which satisfy $a+b+c+d=100$

CantonMathGuy wrote: Find the maximal value of \[S = \sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}},\]where $a$, $b$, $c$, $d$ are nonnegative real numbers which satisfy $a+b+c+d = 100$. In the Israel's last test was the following. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=18.$ Prove that: $$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$

v_Enhance wrote: Also, we use will use the key estimate\begin{align*} 0 &\le \frac{(x-1)^2 (x-7)^2}{x^2+7} \\ &= x^2-16x+71 - \frac{448}{x^2+7} \\ \implies \frac{1}{x^2+7} &\le \frac{x^2-16x+71}{448} \\ \implies \sum_{\text{cyc}} \frac{1}{x^2+7} &\le \sum_{\text{cyc}} \frac{x^2-16x+71}{448} = \frac{6}{7} - \frac{t}{28} \end{align*} can you please say me what is the motivation of considering the identity?

You want the equality to hold at $x=1$ and $x=7$ so that's basically the only possible way to do something like this.

Cvirus213 wrote: v_Enhance wrote: Also, we use will use the key estimate\begin{align*} 0 &\le \frac{(x-1)^2 (x-7)^2}{x^2+7} \\ &= x^2-16x+71 - \frac{448}{x^2+7} \\ \implies \frac{1}{x^2+7} &\le \frac{x^2-16x+71}{448} \\ \implies \sum_{\text{cyc}} \frac{1}{x^2+7} &\le \sum_{\text{cyc}} \frac{x^2-16x+71}{448} = \frac{6}{7} - \frac{t}{28} \end{align*} can you please say me what is the motivation of considering the identity? If you try to do tangent line at $x = 1,7$ you'll find yourself failing because the line through those two points on the graph of $\frac{1}{x^2+7}$ is not tangent to that function at both points; so if you want a polynomial to be tangent at $x = 1,7$ it has to be at least a quadratic.

@above and @2above, thank you very much.

MarkBcc168 wrote: Let $x,y,z,t$ be the permutation of $a,b,c,d$ in ascending order. From rearrangement inequality, we get $$S \leq \sqrt[3]{\frac{x}{t+7}} + \sqrt[3]{\frac{y}{z+7}} + \sqrt[3]{\frac{z}{y+7}} + \sqrt[3]{\frac{t}{x+7}}$$ Why is this true? Is this a different kind of rearrangement inequality than the one here?

CantonMathGuy wrote: Find the maximal value of \[S = \sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}},\]where $a$, $b$, $c$, $d$ are nonnegative real numbers which satisfy $a+b+c+d = 100$. Proposed by Evan Chen, Taiwan Thank you for this beautiful inequality v_enhance! We claim that the maximum value of the given expression is $\boxed{\frac{8}{\sqrt[3]{7}}}$ which is attained for $(a, b, c, d) = (49, 1, 49, 1)$. Here's my proof (similar to many other proofs mention above) : We see that if $T[x, y] = \sqrt[3]{\frac{x}{y+7}}$. We see that $T[x, y] + T[y, x] \leq T[x+y+14, 0]$ by completing Hölder's inequality on $(x + 7)(7 + y)((x+7)+ (y+7))$. Now, for the final showdown, observe that $T[a, b] + T[b, c] + T[c, d] + T[d, a] \leq T[a', d'] + T[d', a'] + T[b', c'] + T[c', b']$ by using Rearrangement inequality on the descending sequences $\left \langle \sqrt[3]{a'} \right \rangle$ and $\left \langle \sqrt[3]{\frac{1}{d'+7}} \right \rangle$ ($a', b', c', d'$ is a permutation of $a, b, c, d$ but in increasing order). and hence we get that $T[a, b] + T[b, c] + T[c, d] + T[d, a] \leq [T, a' + d' +14, 0] + T[b' + c' + 14, 0] \leq$ $ \underset{\mathrm{simple inequality}}{\underline{\sqrt[3]{4\frac{a'+b'+c'+d'}{7}}}} = \boxed{\frac{8}{\sqrt[3]{7}}}$ as desired.

CantonMathGuy wrote: Lemma. [Two-variable version] For all nonnegative real $x$ and $y$, \[ \sqrt[3]{\frac{x}{y+7}} + \sqrt[3]{\frac{y}{x+7}} \le \sqrt[3]{\frac{x+y}{7} + 2}. \] genralzation: For all nonnegative real $x$ and $y$,and positive real$n$ \[ \sqrt[3]{\frac{x}{2y+n}} + \sqrt[3]{\frac{y}{2x+n}} \le \sqrt[3]{\frac{x+y}{n} + 1}.\]an other varient: \[ \sqrt[3]{\frac{nx}{2y+n}} + \sqrt[3]{\frac{ny}{2x+n}} \le \sqrt[3]{x+y+n}.\]

CantonMathGuy wrote: Find the maximal value of \[S = \sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}},\]where $a$, $b$, $c$, $d$ are nonnegative real numbers which satisfy $a+b+c+d = 100$. Proposed by Evan Chen, Taiwan For notation purposes let \((a,b,c,d)\) be that expression. First of all, by re-arrangement the proposed inequality for \((a,b,c,d)\) is true if it is true for \((b,a,d,c)\). So, if \(S\) is the maximum of \[(a,b,c,d)+(b,a,d,c)\]then our desired answer will be \(\frac{S}{2}\). Now this triggers us to find the upper bound of \[\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{a+7}}\]in terms of \(a\) and \(b\). Now the inequality that will ring a bell is Holder, because of the cube root and moreover we want to change this disgusting fractional form into something nice and linear. With that motive, we consider \(\sqrt[3]{\frac{1}{a+7}}\) to be a part of the holder inequality, that is \[\sqrt[3]{7}\sqrt[3]{b}\sqrt[3]{\frac{1}{a+7}}+\sqrt[3]{a}\sqrt[3]{7}\sqrt[3]{\frac{1}{b+7}}\leq \sqrt[3]{(7+a)(7+b)\left(\frac{1}{a+7}+\frac{1}{b+7}\right)}=\sqrt[3]{a+b+14}\]and this seems pretty cute. Now, we need the maximum of \[\sqrt[3]{a+b+14}+\sqrt[3]{b+c+14}+\sqrt[3]{c+d+14}+\sqrt[3]{d+a+14}\]and by holder, it is just \[\sqrt[3]{16(a+b+c+d+a+b+c+d+56)}=\sqrt[3]{2^{12}}=16\]implying that \(S=\frac{16}{\sqrt[3]{7}}\) and so, our desired answer is \(\frac{S}{2}=\frac{8}{\sqrt[3]{7}}\). Now to find the equality, we need \(a=c\) and \(b=d\) and \(a+b=50\). So we need \[\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{a+7}}=\frac{4}{\sqrt[3]{7}}\]and also \(a+b=50\), so \[\sqrt[3]{\frac{50-a}{a+7}}+\sqrt[3]{\frac{a}{57-a}}=\frac{4}{\sqrt[3]{7}}\]Clearly \(a+7\) or \(57-a\) should be a multiple of \(7\), for \(a+7\) we try \(a=0\), and we don't get equality case, for \(57-a\), we try \(a=1\) which indeed gives us the equality case, so the answer is \((49,1,49,1)\) and by symmetry \(a=49\) works too. Now to prove that this is the only possible case, we can consider the graph of the function \[\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{a+7}}=\frac{4}{\sqrt[3]{7}}\]and verify that it intersects the graph in only two possible real points, which are \(1\) and \(49\).

rama1728 wrote: For notation purposes let \((a,b,c,d)\) be that expression. First of all, by re-arrangement the proposed inequality for \((a,b,c,d)\) is true if it is true for \((b,a,d,c)\). So, if \(S\) is the maximum of \[(a,b,c,d)+(b,a,d,c)\]then our desired answer will be \(\frac{S}{2}\). Does this mean that for $a+b+c=3$ the maximum of $a^2b+b^2c+c^2a$ is $\frac{27}{8}$, since we know that $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 \leq \frac{27}{4}$?

MatteD wrote: rama1728 wrote: For notation purposes let \((a,b,c,d)\) be that expression. First of all, by re-arrangement the proposed inequality for \((a,b,c,d)\) is true if it is true for \((b,a,d,c)\). So, if \(S\) is the maximum of \[(a,b,c,d)+(b,a,d,c)\]then our desired answer will be \(\frac{S}{2}\). Does this mean that for $a+b+c=3$ the maximum of $a^2b+b^2c+c^2a$ is $\frac{27}{8}$, since we know that $a^2b+ab^2+b^2c+bc^2+c^2a+ca^2 \leq \frac{27}{4}$? What if we have \(b\geq c\geq a\) wlog?

rama1728 wrote: What if we have \(b\geq c\geq a\) wlog? Then my proof works, if you can show that your condition implies $a^2b+b^2c+c^2a \leq ab^2+bc^2+ca^2$.

MatteD wrote: rama1728 wrote: What if we have \(b\geq c\geq a\) wlog? Then my proof works, if you can show that your condition implies $a^2b+b^2c+c^2a \leq ab^2+bc^2+ca^2$. yeah!

rama1728 wrote: MatteD wrote: rama1728 wrote: What if we have \(b\geq c\geq a\) wlog? Then my proof works, if you can show that your condition implies $a^2b+b^2c+c^2a \leq ab^2+bc^2+ca^2$. yeah! Except this is false, and the maximum is actually $4$. That step in your proof of the problem is wrong, since you would need to prove $(a,b,c,d) \leq (b,a,d,c)$ for all $a,b,c,d$, to prevent something where $(a,b,c,d)$ is greater than $\frac{S}{2}$ but $(a,b,c,d)+(b,a,d,c)\leq S$. You can't show $(a,b,c,d) \leq (b,a,d,c)$ since it's false by symmetry, though. Nevertheless, your proof is not totally hopeless, as you can find in this thread many posts which justify correctly why it is sufficient to consider the quantity $\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{a+7}}$

I get my mistake, thanks. And yeah I checked the posts, instead of me adding it should be me using rearrangement lol

I got spoiled of the "one-liner" of this problem first but didn't initially understand the motivation behind it because it felt so random. Here is some motivation I came up with: First: We start with the inequality $\sqrt[3] {\frac{a}{b+7}}$ $\leq$ $\frac{1}{3} (\frac{a}{a+7} + \frac{a+7}{7} + \frac{7}{b+7})$ which just has the motivation of having $\sum \frac{a}{a+7} + \frac{7}{b+7}$ totaling 4. However, after trying for a while we can't find an equality case - so we want to minimize the right hand side MORE. Note that we can "replace" the term $\frac{a+7}{7}$ with $\frac{a+7}{k}$ for any k $ \in {R+}$, which allows us to make the new inequality $\sqrt[3]{\frac{a}{b+7}}$ $\leq$ $\sum \frac{1}{3} \sqrt[3]{\frac{k}{7}}(\frac{a}{a+7} + \frac{a+7}{k} + \frac{7}{b+7})$ The RHS is minimized when $k^\frac{1}{3} + 32k^\frac{-2}{3}$ is minimized, which occurs when $k = 64$ (which can be confirmed by taking the derivative of $k^\frac{1}{3} + 32k^\frac{-2}{3}$ or by weighted AM-GM) We can then plug in $k = 64$ into our original inequality to find the one-liner itself: $\sum \sqrt[3]{\frac{64a}{b+7}}$ $\leq$ $\sum \frac{1}{3}(\frac{a+7}{64} + \frac{a}{a+7} + \frac{7}{b})$ = $\frac{8}{\sqrt[3]{7}}$ This holds when we set $a = c = 1$ or $49$, $b = d = 50-a$ Sorry for poor formatting, still kind of bad at latexing

Let $w,x,y,z$ be a permutation of $a,b,c,d$ s.t. $w\leq x\leq y\leq z.$ By Rearrangement, Holder and AM-GM successively we have: $$\sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}}$$$$ \leq \sqrt[3]{\frac{w}{z+7}} +\sqrt[3]{\frac{x}{y+7}} +\sqrt[3]{\frac{y}{x+7}} +\sqrt[3]{\frac{z}{w+7}}$$$$\leq \sqrt[3]{\frac{w+z+14}{7}} +\sqrt[3]{\frac{x+y+14}{7}}$$$$\leq 2\sqrt[3]{\frac{w+x+y+z+28}{14}} =\sqrt[3]{\frac{512}{7}}$$

ZETA_in_olympiad wrote: WLOG $a\leq b\leq c\leq d.$ This inequality is only cyclic, not symmetric, so you can't do this. The best you can say is that $a = \min(a,b,c,d)$.

djmathman wrote: ZETA_in_olympiad wrote: WLOG $a\leq b\leq c\leq d.$ This inequality is only cyclic, not symmetric, so you can't do this. The best you can say is that $a = \min(a,b,c,d)$. You are right. This should work. Let $w,x,y,z$ be a permutation of $a,b,c,d$ s.t. $w\leq x\leq y\leq z.$ Now I just have to change variables from second ļine.

Oops I used desmos in the last step. Let $a_1+a_2+a_3+a_4=100$ If $a_1\ge a_2\ge a_3\ge a_4$, for any permutation $\pi$ of $\{1,2,3,4\}$, $$\sum_{i=1}^4 \left( \frac{a_i}{a_{\pi(i)}+7}\right)^{1/3} \le \sum_{i=1}^4 \left( \frac{a_i}{a_{5-i}+7}\right)^{1/3}$$ Lemma: if $x+y=S$, $(x/(y+7))^{\frac 13}+(y/(x+7))^{\frac 13}$ is maximized if $xy=49$ Proof: first derivative bash to get $(2x-S)(x^2-Sx+49)=0$ Now we let $$g(S)=(\frac{S-\sqrt{S^2-196}}{S+\sqrt{S^2-196}+14})^{\frac 13} + (\frac{S+\sqrt{S^2-196}}{S-\sqrt{S^2-196}+14})^{\frac 13} $$ $g''(S)<0$ for all $S\ge 14$ so we maximize $g(S)+g(100-S)$ and get the equality case is $(49,1,49,1)$

Was spoiled knowing the equality case going in oops, probably couldn't have solved it without knowing that We claim that the answer is $\sqrt[3]{512/7}$, achieved by $(a,b,c,d)=(49,1,49,1).$ First, let $p,q,r,s$ such that $p\leq q\leq r\leq s$, and the multiset $\{p,q,r,s\}$ is the same as the multiset $\{a,b,c,d\}$. Then, by Rearrangement Inequality on $(\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c},\sqrt[3]{d})$ and $(\sqrt[3]{1/(a+7)},\sqrt[3]{1/(b+7)},\sqrt[3]{1/(c+7)},\sqrt[3]{1/(d+7)})$ $$S\leq \sqrt[3]{\frac{p}{s+7}}+\sqrt[3]{\frac{q}{r+7}}+\sqrt[3]{\frac{r}{q+7}}+\sqrt[3]{\frac{s}{p+7}}.$$ Main Claim: $$\sqrt[3]{\frac{p}{s+7}}+\sqrt[3]{\frac{s}{p+7}}\leq \sqrt[3]{\frac{p+s}{7}+2}.$$By Holder, $$p^{1/3}(\frac{1}{s+7})^{1/3}7^{1/3}+s^{1/3}(\frac{1}{p+7})^{1/3}7^{1/3}\geq \sqrt[3]{(p+7)(s+7)(\frac{1}{s+7}+\frac{1}{p+7})}=7^{1/3}\sqrt[3]{\frac{p+s}{7}+2},$$which shows the claim. (The Holder application, particularly the inclusion of the constant factor, was motivated by the equality case) Thus, $$S\leq \sqrt[3]{\frac{p+s}{7}+2}+\sqrt[3]{\frac{q+r}{7}+2}.$$ We are almost done. Since cube roots are concave, by Jensen's $$\sqrt[3]{\frac{p+s}{7}+2}+\sqrt[3]{\frac{q+r}{7}+2}\leq 2\sqrt[3]{2+\frac{p+q+r+s}{14}}=\sqrt[3]{512/7},$$as desired.

Let us solve the two-variable verion first. We will show that \[\sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{a+7}}\le \sqrt[3]{\frac{a+b+14}{7}}\]Note that by Holder's inequality we have that \[\left(\sqrt[3]{7}\sqrt[3]{a}\sqrt[3]{\frac{1}{b+7}}+\sqrt[3]{7}\sqrt[3]{b}\sqrt[3]{\frac{1}{a+7}}\right)^3\le (a+7)(b+7)\left(\frac{1}{a+7}+\frac{1}{b+7}\right)\]which proves our claim. Note that we can now do the following: let $p>q>r>s$ but $p,q,r,s$ is a permutation of $a,b,c,d$. Note that by rearrangement inequality, \[S\le \sqrt[3]{\frac{p}{s+7}} + \sqrt[3]{\frac{s}{p+7}} + \sqrt[3]{\frac{q}{r+7}} + \sqrt[3]{\frac{s}{r+7}}\le \sqrt[3]{\frac{p+s+14}{7}}+\sqrt[3]{\frac{q+r+14}{7}}\]Now, the rest is easy. We can do the following: let $\tfrac{p+s+14}{7}=x$ and $\frac{q+r+14}{7}=y$ and $x+y=\tfrac{128}{7}$ then \[\sqrt[3]{x}+\sqrt[3]{y}\le 2\sqrt[3]{\frac{x+y}{2}}=\frac{8}{\sqrt[3]{7}}\]

a1267ab wrote: v_Enhance wrote: I'll let someone else post the one-liner. If you insist. \[\sqrt[3]{\frac{7a}{64(b+7)}}\leq \frac{1}{3}\left(\frac{a+7}{64}+\frac{a}{a+7}+\frac{7}{b+7}\right).\quad \blacksquare\]Attributed to "Russian jury". Interestingly enough, I've found similar techniques used on other contest problems (including this one, which ironically is sourced as ``Russian olympiad"): $$\sqrt[3]{(a+b+c)(a+b+d)} \geq \sqrt[3]{ac} +\sqrt[3]{bd}.$$This can be done by summing \begin{align*} \frac a{a+b} + \frac c{a+b+c} + \frac{a+b}{a+b+d} &\geq 3\sqrt[3]{\frac{ac}{(a+b+c)(a+b+d)}} \\ \frac b{a+b} + \frac{a+b}{a+b+c} + \frac d{a+b+d} &\geq 3\sqrt[3]{\frac{bd}{(a+b+c)(a+b+d)}}. \end{align*}

HamstPan38825 wrote: Interestingly enough, I've found similar techniques used on other contest problems That might be because any time Hölder is used it can also be replaced by an appropriate sum of AM-GM inequalities (since you can prove Hölder by summing AM-GM's).

I claim the answer is $(\frac{512}{7})^\frac{1}{3}$, which is achievable by $(49,1,49,49)$. By rearrangement inequality, it suffices to check $$a^\frac{1}{3}(b+7)^\frac{-1}{3}+b^\frac{1}{3}(a+7)^\frac{-1}{3}+c^\frac{1}{3}(d+7)^\frac{-1}{3}+d^\frac{1}{3}(c+7)^\frac{-1}{3} \le (\frac{512}{7})^\frac{1}{3}$$where $a+b+c+d=100$. For now just work with maximizing $a^\frac{1}{3}(b+7)^\frac{-1}{3}+b^\frac{1}{3}(a+7)^\frac{-1}{3}$ subject to $a+b=x$. Taking the derivative here, after a bit of computations, we find that the derivative is $0$ only when $(2a-x)(a^2-ax+49)=0$. This means our possible maxima satisfy either $ab=0$, $a=\frac{x}{2}$, or $ab=49$. We can easily check the second derivative when $a=\frac{x}{2}$ is positive. We can also check in the $ab=49$ case, then subsituting $a=7y$, $b=\frac{7}{y}$, that our thing simplifies to $(\frac{(y+1)^2}{y})^\frac{1}{3}$, which is equivalent to $(2+\frac{1}{7}x)^\frac{1}{3}$. We can easily check this is bigger than what we get if $a=0$. This means our big expression from earlier is at most $$(2+\frac{1}{7}x)^\frac{1}{3}+(2+\frac{1}{7}y)^\frac{1}{3},$$subject to $x+y=100$. From here, we get by Jensen that this is maximized at $x=y=50$, which gives our equality case (and bound) from earlier. Edit: Yay! I finished A shortlist! Onto C shortlist.

The original problem is a special case of the following generalization where $$\beta=100,\lambda=7$$.

Generalization 1 Let $a,b,c,d$ be nonnegative reals such that $a+b+c+d=\beta$. Then prove that $$\sqrt[3]{\dfrac{a}{b+\lambda}}+\sqrt[3]{\dfrac{b}{c+\lambda}}+\sqrt[3]{\dfrac{c}{d+\lambda}}+\sqrt[3]{\dfrac{d}{a+\lambda}}\leq \sqrt[3]{\dfrac{4\left(\beta + 4\lambda\right)}{\lambda}}$$

ehuseyinyigit wrote: Generalization 1 Let $a,b,c,d$ be nonnegative reals such that $a+b+c+d=\beta$. Then prove that $$\sqrt[3]{\dfrac{a}{b+\lambda}}+\sqrt[3]{\dfrac{b}{c+\lambda}}+\sqrt[3]{\dfrac{c}{d+\lambda}}+\sqrt[3]{\dfrac{d}{a+\lambda}}\leq \sqrt[3]{\dfrac{4\left(\beta + 4\lambda\right)}{\lambda}}$$ By Holder's inequality, $$\sum_{\text{cyc}} \sqrt[3]{\frac{a}{b+\lambda}} \leq \sqrt[3]{\sum_{\text{cyc}} a+\lambda}\sqrt[3]{\sum_{\text{cyc}}\frac{a}{a+\lambda}}\sqrt[3]{\sum_{\text{cyc}} \frac{1}{b+\lambda}} =\sqrt[3]{\left(\beta+4\lambda\right)\left(\sum_{\text{cyc}}\frac{a}{a+\lambda}\right)\left(\sum_{\text{cyc}} \frac{1}{b+\lambda}\right)}$$ Also, notice that by AM-GM inequality, $$2\sqrt{\left(\sum_{\text{cyc}} \frac{a}{a+\lambda}\right)\left(\sum_{\text{cyc}} \frac{\lambda}{b+\lambda}\right)} \leq 4 \implies \sqrt[3]{\left(\sum_{\text{cyc}}\frac{a}{a+\lambda}\right)\left(\sum_{\text{cyc}} \frac{1}{b+\lambda}\right)}\leq \sqrt[3]{\frac{4}{\lambda}}.$$ And hence, we get $\sum_{\text{cyc}} \sqrt[3]{\frac{a}{b+\lambda}}\leq\sqrt[3]{\dfrac{4\left(\beta + 4\lambda\right)}{\lambda}}$, as required. $\square$

What about this one conjectured again by me ?