$$\sum \frac{1}{a^3+b+c+d} \le \sum \frac{1}{4 \sqrt{a}}$$So we should prove that $$\sum \frac{1}{\sqrt{a}} \le a+b+c+d$$If we say $a= \frac{x}{y} \;\; b= \frac{y}{z} \;\; c=\frac{z}{t} \;\; d=\frac{t}{x}$ and arrange the given inequality we have $$\sum x^2zt \ge \sum x^{\frac{1}{2}}y^{\frac{3}{2}}zt $$which is true by Muirhead.

electrovector wrote: $$\sum \frac{1}{a^3+b+c+d} \le \sum \frac{1}{4 \sqrt{a}}$$So we should prove that $$\sum \frac{1}{\sqrt{a}} \le a+b+c+d$$If we say $a= \frac{x}{y} \;\; b= \frac{y}{z} \;\; c=\frac{z}{t} \;\; d=\frac{t}{x}$ and arrange the given inequality we have $$\sum x^2zt \ge \sum x^{\frac{1}{2}}y^{\frac{3}{2}}zt $$which is true by Muirhead. $$\sum x^2zt \ge \sum x^{\frac{1}{2}}y^{\frac{3}{2}}zt $$is equivalent $(2,1,1,0) \ge (\frac{3}{2},1,1,\frac{1}{2})$????

Jjesus wrote: $$\sum x^2zt \ge \sum x^{\frac{1}{2}}y^{\frac{3}{2}}zt $$is equivalent $(2,1,1,0) \ge (\frac{3}{2},1,1,\frac{1}{2})$???? Equivalent to $(2, 0, 1, 1) \ge (\frac{1}{2}, \frac{3}{2}, 1, 1)$

electrovector wrote: Jjesus wrote: $$\sum x^2zt \ge \sum x^{\frac{1}{2}}y^{\frac{3}{2}}zt $$is equivalent $(2,1,1,0) \ge (\frac{3}{2},1,1,\frac{1}{2})$???? Equivalent to $(2, 0, 1, 1) \ge (\frac{1}{2}, \frac{3}{2}, 1, 1)$ $$\sum \frac{1}{\sqrt{a}} \le a+b+c+d$$$$ x^{\frac{1}{2}}y^{\frac{3}{2}}zt+y^{\frac{1}{2}}z^{\frac{3}{2}}tx+z^{\frac{1}{2}}t^{\frac{3}{2}}xy+t^{\frac{1}{2}}x^{\frac{3}{2}}yz \le x^2zt+y^2tx+z^2xy+t^2yz$$and $(2,0,1,1)=x^2yz+x^2yt+x^2tz+y^2xt+y^2xz+y^2zt+z^2xy+z^2xt+z^2yt+t^2xy+t^2xz+t^2yz$

Redacted I am so bad at inequality..

Afo wrote: By AM-GM: $$\sum_{cyc} a^3+b+c+d \ge \sum_{cyc} 4\sqrt[4]{a^2abcd} = \sum_{cyc} 4\sqrt{a}\iff \sum_{cyc} \frac{1}{a^3+b+c+d} \leq\sum_{cyc} \frac{1}{4\sqrt{a}}$$ Letting $\sqrt{a}=p,\sqrt{b}=q,\sqrt{c}=r,\sqrt{d}=s$. And so $abcd=1\implies pqrs=1$. The question is equivalent to proving $$p^2+q^2+r^2+s^2\ge \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s} = pqr+pqs+prs+qrs$$ Which is true by AM-GM on $p^2+q^2+r^2 \ge 3pqr$ and summing cyclically and dividing by three. I don't think that $p^2+q^2+r^2 \geq 3pqr$ is true by AM-GM. $p^2+q^2+r^2 \geq \sqrt[3]{(pqr)^2}$ is true.

Steve12345 wrote: Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Prove the inequality $\frac{1}{a^3 + b + c + d} +\frac{1}{a + b^3 + c + d}+\frac{1}{a + b + c^3 + d} +\frac{1}{a + b + c + d^3} \leq \frac{a+b+c+d}{4}$ Proposed by Romania This is my problem. We didn't know it was in SHL. Thank you for the information.

Steve12345 wrote: Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Prove the inequality $\frac{1}{a^3 + b + c + d} +\frac{1}{a + b^3 + c + d}+\frac{1}{a + b + c^3 + d} +\frac{1}{a + b + c + d^3} \leq \frac{a+b+c+d}{4}$ Proposed by Romania Hello friends! Is It OK if I multiply LHS by $abcd$, and then due to homogeneity assume that $a+b+c+d=1$? Or am I wrong? Thank you! CROWmatician

$$abc+abd+acd+bcd\leq a^2+b^2+c^2+d^2$$it's false, and so $$\sum\frac{1}{\sqrt a}\leq a+b+c+d$$it's false in the given problem.

I gave solution to wrong problem

CROWmatician wrote: Hello friends! Is It OK if I multiply LHS by $abcd$, and then due to homogeneity assume that $a+b+c+d=1$? Or am I wrong? Thank you! CROWmatician $$\frac{abcd}{a^3 + b + c + d} +\frac{abcd}{a + b^3 + c + d}+\frac{abcd}{a + b + c^3 + d} +\frac{abcd}{a + b + c + d^3} \leq \frac{a+b+c+d}{4}$$is not homogenous. Sorry, you can't.

electrovector wrote: $$\sum x^2zt \ge \sum x^{\frac{1}{2}}y^{\frac{3}{2}}zt $$which is true by Muirhead. Muirhead works for symmetric inequalities, not cyclic.

Good. It's exactly my solution (and the official solution. I think).

Since $abcd=1$ involves $a+b+c+d\ge 4$, it suffices to prove the inequality for $a+b+c+d\ge 4$. Replacing $a,b,c,d$ by $ka$, $kb$, $kc$, $kd$ with $a+b+c+d=4$, it follows immediately that it suffices to consider the case $k=1$, that is $a+b+c+d=4$. Actually, the following inequality holds: $$\frac{1}{a_1^k + a_2 + \cdots + a_n} +\frac{1}{a_1 + a_2^k +\cdots + a_n}+\cdots+\frac{1}{a_1 + a_2 +\cdots+ a_n^k }\leq 1$$for $a_1+a_2+\cdots+a_n\ge n$ and $k>1$ (see Cirtoaje's book Mathematical Inequalities, vol. 4, P 3.12, page 242 - http://ace.upg-ploiesti.ro/vcirtoaje/vcirtoaje.php). PS. For $k>1$ and $a_1+a_2+\cdots+a_n\ge n$ (therefore for $a_1a_2\cdots a_n=1$), the inequality is true: $$\frac{1}{a_1^k + a_2 + \cdots + a_n} +\frac{1}{a_1 + a_2^k +\cdots + a_n}+\cdots+\frac{1}{a_1 + a_2 +\cdots+ a_n^k }\leq \frac n{a_1+a_2+\cdots+a_n}.$$

Steve12345 wrote: Let $a, b, c, d$ be positive real numbers such that $abcd = 1$. Prove the inequality $\frac{1}{a^3 + b + c + d} +\frac{1}{a + b^3 + c + d}+\frac{1}{a + b + c^3 + d} +\frac{1}{a + b + c + d^3} \leq \frac{a+b+c+d}{4}$ Proposed by Romania We have 2 claims, Claim 1: $\frac 34 \cdot (a+b+c+d)^3 \ge 12(a+b+c+d)$ Proof: By AM-GM, $a+b+c+d \ge 4$, so $$\frac 34 \cdot (a+b+c+d)^3 =\frac 34 \cdot (a+b+c+d)^2(a+b+c+d) \ge \frac 34 \cdot 16(a+b+c+d) =12(a+b+c+d)$$. Hence our claim is true. Claim 2:$\frac {(a+b+c+d)^3}{4} \ge 4\sum abc$ Proof: $$\frac {(a+b+c+d)^3}{4} \ge 4\sum abc \iff (a+b+c+d)^3 \ge 16\sum abc \iff \sum a^3 +3\sum a^2b \ge 10\sum abc$$which is obvious by Muirhead. Hence this claim is also true. Now, By cauchy $$(a^3+b+c+d)\left(\frac 1a +b+c+d\right)=(a^3+b+c+d)(bcd+b+c+d) \ge (a+b+c+d)^2 \implies \frac {1}{a^3+b+c+d} \le \frac {bcd+b+c+d}{(a+b+c+d)^2}$$on summing cyclically, we want to prove $$\frac {a+b+c+d}{4} \ge \frac {\sum abc +3(a+b+c+d)}{(a+b+c+d)^2} \iff (a+b+c+d)^3 \ge 4\sum abc +12(a+b+c+d)$$which follows by summing the 2 claims. $Q.E.D$

Very beautiful.

Similar to the above solutions, but in a more motivated fashion. We try to get upper bounds on the denominator instead of going for a lower bound directly. Using Cauchy, we get that $(a^3+b+c+d)(\frac{1}{a}+b+c+d)\geq (a+b+c+d)^2$(don't use AM-GM to get the $16$ bound yet, keep the term until you have a certain path). Rearranging gives that $\frac{1}{a^3+b+c+d}\leq \frac{\frac{1}{a}+b+c+d}{(a+b+c+d)^2}$. Similar cyclic things hold, so we get that we are just trying to prove that $\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+3(a+b+c+d)}{(a+b+c+d)^2}\leq \frac{a+b+c+d}{4}$. Rearranging gives that $4(\frac{1}{a}+\frac 1 b+\frac 1 c+\frac 1 d+3(a+b+c+d))\leq (a+b+c+d)^3$. To try to make the inequality nicer, we convert the $\frac{1}{a}$ to a $bcd$ and similarly for the others, to get that we must prove that $4(abc+abd+acd+bcd)+12(a+b+c+d)\leq (a+b+c+d)^3$. The following claim is pretty famous apparently, but we'll present a proof for completeness: Claim: $4(abc+abd+acd+bcd)\leq \frac{(a+b+c+d)^3}{4}$. Pf: This basically just reduces to showing that $\sum a^3 + 3\sum a^2b+6\sum abc\geq 16\sum abc\implies \sum a^3+3\sum a^2b\geq 10\sum abc$. This readily follows from muirhead, because $(3,0,0,0)$ > $(1,1,1,0)$ gives $6(a^3+b^3+c^3+d^3)\geq 6(abc+abd+acd+bcd)\implies a^3+b^3+c^3+d^3\geq abc+abd+acd+bcd$, and so all we have to prove is that $\sum a^2b\geq 3\sum abc,$ which also follows from muirhead because $(2,1,0,0)$ > $(1,1,1,0)$ gives the desired inequality, with both sides doubled. The claim is proven. We now finish. We have that $4(abc+abd+acd+bcd)+12(a+b+c+d)\leq 12(a+b+c+d)+\frac{(a+b+c+d)^3}{4},$ so we want to show that $12(a+b+c+d)\leq \frac{3}{4}(a+b+c+d)^3\implies (a+b+c+d)^2\geq 16,$ which easily follow from AM-GM. Done.

csav10 wrote: Since $abcd=1$ involves $a+b+c+d\ge 4$, it suffices to prove the inequality for $a+b+c+d\ge 4$. Replacing $a,b,c,d$ by $ka$, $kb$, $kc$, $kd$ with $a+b+c+d=4$, it follows immediately that it suffices to consider the case $k=1$, that is $a+b+c+d=4$. Actually, the following inequality holds: $$\frac{1}{a_1^k + a_2 + \cdots + a_n} +\frac{1}{a_1 + a_2^k +\cdots + a_n}+\cdots+\frac{1}{a_1 + a_2 +\cdots+ a_n^k }\leq 1$$for $a_1+a_2+\cdots+a_n\ge n$ and $k>1$ (see Cirtoaje's book Mathematical Inequalities, vol. 4, P 3.12, page 242 - http://ace.upg-ploiesti.ro/vcirtoaje/vcirtoaje.php). PS. For $k>1$ and $a_1+a_2+\cdots+a_n\ge n$ (therefore for $a_1a_2\cdots a_n=1$), the inequality is true: $$\frac{1}{a_1^k + a_2 + \cdots + a_n} +\frac{1}{a_1 + a_2^k +\cdots + a_n}+\cdots+\frac{1}{a_1 + a_2 +\cdots+ a_n^k }\leq \frac n{a_1+a_2+\cdots+a_n}.$$ I think I found a lagrange multipler solution to this, can someone check this is actually right and there are no mistakes? ( If there are, please mention it, I am trying to learn lagrange mutliplers )

The same thing can be done to the original JBMO question

\[\sum{\frac{1}{(a+b+c+d)(a^3+b+c+d)}\leq \sum{\frac{1}{(a^2+b+c+d)^2}}}\leq \sum{\frac{(1+b+c+d)^2}{(a+b+c+d)^4}}\overset{?}{\leq} \frac{1}{4}\]\[\sum{(1+b+c+d)^2}\overset{?}{\leq}4(\sum{a})^2\leq \frac{(\sum{a})^4}{4}\iff 4\sum{a^2}+8\sum{ab}\overset{?}{\geq}3\sum{a^2}+4\sum{ab}+6\sum{a}+4\]\[\sum{a^2}+4\sum{ab}\overset{?}{\geq} 6\sum{a}+4\]Which is true since \[\sum{a^2}+4\sum{ab}-6\sum{a}-4\geq (\sum{a})^2+12-6\sum{a}-4=(\sum{a})^2-6\sum{a}+8=((\sum{a})-4)((\sum{a})-2)\geq 0\]As desired.$\blacksquare$

I discussed this problem in my youtube channel (little fermat) Video in my inequalities tutorial playlist.