EDIT: Cyclic Squares has a video on the motivation behind this solution now: https://www.youtube.com/watch?v=LSYP_KMbBNc

... and it really happened. Fortunately, it's solved by the tangent line trick. We observe the miraculous identity \[ \frac{1}{b^3+4} \ge \frac14 - \frac{b}{12} \]since $12-(3-b)(b^3+4) = b(b+1)(b-2)^2 \ge 0$. Moreover, \[ ab+bc+cd+da = (a+c)(b+d) \le \left( \frac{(a+c)+(b+d)}{2} \right)^2 = 4. \]Thus \[ \sum_{\text{cyc}} \frac{a}{b^3+4} \ge \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \ge 1 - \frac13 = \frac23. \]This minimum $\frac23$ is achieved at $(a,b,c,d) = (2,2,0,0)$ and permutations.

...tfw you think $(1,1,1,1)$ is the minimum case...

lucasxia01 wrote:

Really? I tried that a little but couldn't see any solution by Jensen's.

jasonhu4 wrote: ...tfw you think $(1,1,1,1)$ is the minimum case... That was me

r3mark wrote: lucasxia01 wrote:

Really? I tried that a little but couldn't see any solution by Jensen's. He's trolling, the function isn't convex.

Why is there no meme yet about 3 variable vs 4 variable inequalities

r3mark wrote: lucasxia01 wrote:

Really? I tried that a little but couldn't see any solution by Jensen's. It's a reference to a some meme.

How many points would you get for correctly guessing the minimum and finding the equality case?

Plasma_Vortex wrote: How many points would you get for correctly guessing the minimum and finding the equality case? I'd say 1

bobthesmartypants wrote: Plasma_Vortex wrote: How many points would you get for correctly guessing the minimum and finding the equality case? I'd say 1 I doubt it gets any points, that kind of work is basically a leap of faith guess with no proving involved.

bobthesmartypants wrote: Plasma_Vortex wrote: How many points would you get for correctly guessing the minimum and finding the equality case? I'd say 1 Really? I did that and showed that it was at least a local minimum when $c=d=0.$ I also wrote down a short proof for the maximum being 1 cuz I had time left.

whatshisbucket wrote: bobthesmartypants wrote: Plasma_Vortex wrote: How many points would you get for correctly guessing the minimum and finding the equality case? I'd say 1 Really? I did that and showed that it was at least a local minimum when $c=d=0.$ I also wrote down a short proof for the maximum being 1 cuz I had time left. I did the same thing, but I proved that it was the global minimum in [0,4] (which is true). Would that get 2 points?

This problem is somewhat similar to CGMO 2007 P3, which I recall was on a Red MOP handout in 2016.

Is it also true that $$\frac{a}{b^3+4} + \frac{b}{c^3+4} + \frac{c}{d^3+4} + \frac{d}{a^3+4} \ge \min\left\{\frac{a+c}{(b+d)^3+4} + \frac{b+d}{4}, \frac{b+d}{(a+c)^3+4} + \frac{a+c}{4}\right\} \ \ ?$$ edit: no, it's false..

How many points would I get for claiming $\frac{2}{3}$ is the answer and drawing a volcano called "Mt. Inequality" erupting?

This should have been JMO #6 instead

My solution: Observe that $\frac{1}{b^3+4} \ge - \frac{b}{12} + \frac{1}{4}$, which is true upon factorising. Also by AM-GM, $ab+bc+cd+da \le 4$. So the answer is $\frac{2}{3}$, and equality holds if 2 of $a,b,c,d$ are $2$ and the other two are $0$.

Who wants the good old days where AM-GM inequalities were on the IMO

Generalization 2 Let $a,b,c,d,k$ be nonnegative reels such that $a+b+c+d=k^k$. Then prove that $$\dfrac{a}{b^{k+1}+k^k}+\dfrac{b}{c^{k+1}+k^k}+\dfrac{c}{d^{k+1}+k^k}+\dfrac{d}{d^{k+1}+k^k}\geq \dfrac{k(4-k^{k-1})+4}{4k+4}$$

Generalization 3 Let $a,b,c,d,n,k$ be nonnegative reels such that $n>0$ and $a+b+c+d=n^k$. Then prove that $$\dfrac{a}{\dfrac{kb^{k+1}}{n}+n^k}+\dfrac{b}{\dfrac{kc^{k+1}}{n}+n^k}+\dfrac{c}{\dfrac{kd^{k+1}}{n}+n^k}+\dfrac{d}{\dfrac{ka^{k+1}}{n}+n^k}\geq \dfrac{k(4-n^{k-1})+4}{4k+4}$$

Generalization- USAMO 2017 #6

ehuseyinyigit wrote: Generalization 3 Let $a,b,c,d,n,k$ be nonnegative reels such that $n>0$ and $a+b+c+d=n^k$. Then prove that $$\dfrac{a}{\dfrac{kb^{k+1}}{n}+n^k}+\dfrac{b}{\dfrac{kc^{k+1}}{n}+n^k}+\dfrac{c}{\dfrac{kd^{k+1}}{n}+n^k}+\dfrac{d}{\dfrac{ka^{k+1}}{n}+n^k}\geq \dfrac{k(4-n^{k-1})+4}{4k+4}$$ By giving $n=k=2$, we get the original problem

Generalization 4 Let $a_{1},a_{2},\cdots,a_{p},n,k$ be non negative reels, $n>0, p\geq 3$ and $\sum_{cyc}{a_{1}}=n^k$. Then prove that $$\sum_{cyc}{\dfrac{a_{1}}{\dfrac{ka_{2}^{k+1}}{n}+n^k}}\geq \dfrac{k(4-n^{k-1})+4}{4k+4}$$

We claim the answer is $\boxed{\frac 23}$. This lower bound is attainable with $(2,2,0,0)$ Let $f(x) = \frac{1}{x^3+4}$. From the Tangent Line Trick, we claim \[f(x) \ge f(2) + (x-2) \cdot f'(2) = - \frac{1}{12} x + \frac 14,\] which we can prove by expanding. Thus our expression can be written as \begin{align*} \sum_{\text{cyc}} a \cdot f(b) &\ge - \frac{1}{12} \sum_{\text{cyc}} ab + \frac 14 \sum_{\text{cyc}} a \\ &= - \frac{1}{12} (a+c)(b+d) + \frac 14 (a+b+c+d) \\ &\ge - \frac{1}{12} \left(\frac{a+b+c+d}{2}\right)^2 + \frac 14 (a+b+c+d) \\ &= \frac 23.~\blacksquare \end{align*}

We claim the answer is $\frac{2}{3}$ achieved at $(2, 2, 0, 0)$. Now we will prove the bound. Define $f(x)= \frac{1}{x^3+4}$. We will show, \begin{align*} f(x) &\geq f(2) + (x-2) \cdot f'(2)\\ \frac{1}{x^3+4} &\geq \frac{1}{12} + (2-x) \cdot \frac{1}{12}\\ 12 &\geq x^3+4 + (2-x)(x^3+4)\\ (x-2)(x^3+4) - x^3 + 8 &\geq 0\\ x^4 + 4x - 3x^3 &\geq 0\\ x(x^3 -3x^2 + 4) &\geq 0\\ x(x+1)(x-2)^2 &\geq 0\\ \end{align*}as desired. Now using this tangent-line approximation of the function we find, \begin{align*} \sum_{cyc} \frac{a}{b^3+4} \geq \sum_{cyc} a \left(\frac{1}{4} - \frac{b}{12} \right) \end{align*}Now rearranging the right hand side we find, \begin{align*} \sum_{cyc} a \left(\frac{1}{4} - \frac{b}{12} \right) \geq 1 - \frac{ab + bc + cd + da}{12} &= 1 - \frac{(a+c)(b+d)}{12}\\ &\geq 1 - \frac{4}{12} = \frac{2}{3} \end{align*}as desired.

First, one can notice that (Tangent line trick at 2): $$ x(x+1)(x-2)^2 \ge 0 \iff \frac{1}{x^3+4} \ge \frac14 - \frac{x}{12} $$ Thus : $$S \geq 1-\frac{(a+c)(b+d)}{12} \geq 1-\frac{(a+c+b+d)^2}{4 \times 12} = \frac{2}{3}$$ Equality can be achieved when $a=b=2$ and $c=d=0$. $$\mathbb{Q.E.D.}$$

If $f(x)=\frac1{x^3+4}-\frac14+\frac x{12}$ then we have $f(0)=f(2)=0$ and $f'(x)$ is positive for $x>2$ and is positive for small $x$ and has only one root between $0$ and $2$ so by rolles theorem and stuff $f(x)$ is nonnegative on all nonnegatives. Thus we have \[\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4} \ge \frac{a+b+c+d}4-\frac{ab+bc+cd+da}{12}=1-\frac{(a+c)(b+d)}{12}\ge 1-\frac{4}{12}=\frac 23\]by AM-GM with equality when $(a,b,c,d)=(2,2,0,0)$ or cyclic permutations.

omar1tun wrote: hi can any one plz now a site were i can learn Olympiads from beginners level to pro your on that site :skull:

Let $f(x)=\frac{1}{x^3+4}$ Let $g(x)=f(x)-(-\frac{x}{12}+\frac{1}{4})$ Then $g'(x)=-\frac{3x^2}{(x^3+4)^2}+\frac{1}{12}=\frac{(x-2)(x^2+2x-2)(x^3+6x+4)}{12(x^3+4)^2}$. The real zeroes of this are $2$ and $-1 \pm \sqrt{3}$. Plugging them in gives that from $-1-\sqrt{3}$ to $-1+\sqrt{3}$, it increases. Then, from $-1+\sqrt{3}$ to $2$ it decreases. Finally, it increases. Because $g(0)=0$, the first increasing section is always nonnegative. Because $g(2)=0$, the first decreasing part is always nonnegative. So the last part would also be nonnegative beacuse it is more than $g(2)=0$. So $g(x) \ge 0$ $f(x) \ge -\frac{x}{12}+\frac{1}{4}$ $af(b)+bf(c)+cf(d)+df(a) \ge -\frac{ab+bc+cd+da}{12}+\frac{a+b+c+d}{4} \ge -\frac{(b+d)(a+c)}{12}+1 \ge -\frac{2*2}{12}+1 \ge \boxed{\frac{2}{3}}$

We claim that the maximum value is $\boxed{\frac{2}{3}}$. This is given by $(a, b, c, d) = (2, 2, 0, 0)$ and permutations. We claim that \[\frac{1}{a^3+4} \geq \frac{-a+3}{12}\]for $a \in [0, 4]$. Mutliplying both sides by $12a^3+48$ we get \[b^4-3b^3+4b \geq 0 \Rightarrow b(b+1)(b-2)^2 \geq 0\]which is obviously true for $a \in [0, 4]$. Now we have transformed the given to \[\sum_{\text{cyc}} \frac{3a-ab}{12} \geq \frac{2}{3} \Rightarrow \sum_{\text{cyc}} \frac{-ab}{12} \geq \frac{-1}{3} \Rightarrow \sum_{\text{cyc}} ab \leq 4\]This last inequality is true because \[\sum_{\text{cyc}} ab = (a+c)(b+d) = (2-x)(2+x) \leq 4\]$\blacksquare$

For $(a,b,c,d) = (2,2,0,0)$, we take $\displaystyle\frac{2}{3}$. We will prove that this is the minimum. Using the tangent-line trick, we get \[\frac{1}{b^3+4} \geq \frac{1}{4} - \frac{b}{12}.\] Also, from AM-GM, we have \[(a+c)(b+d) \leq \left(\frac{a+b+c+d}{2}\right)^2 = 4.\] So we have \[\sum_{cyc}\frac{a}{b^3+4} \geq \frac{a+b+c+d}{4} - \frac{ab+bc+cd+da}{12} \geq 1 - \frac{1}{3} = \frac{2}{3}.\] Equality $(a,b,c,d) = (2,2,0,0)$ and transfers.

By tangent line trick we can see that $$\frac{1}{x^3+4} \geq \frac{1}{4}-\frac{x}{12}x$$for nonnegative $x,$ with equality at $x=0,2.$ Next note that $(a+c)(b+d) \leq \frac{(a+b+c+d)^2}{4}=4.$ Therefore the original expression is $$\geq \frac{a+b+c+d}{4}-\frac{4}{12}=\boxed{\frac23},$$which is achievable at $(2,2,0,0).\blacksquare$

you guys are still bumping this?