Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in $\LaTeX$. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

This promblem seems to be very difficult and beautiful!!! Does anybody know some solution???

answer: 3/32 is best possible. proof: let v1,v2,...,vn be vertices of graph g and E be set of its edges.for i=1,2,...,nlet A i be set of vertices connectedto vi by an edge,Gi subgraph of g whose set of vertices is AiandEi set of edges of Gi.aNd vi,ei,ti=f(Gi) be no. of vertices,edges and tri. in Gi res. note,no. of tetrashedra and tri. one of whose vertices is Vi are res. ti,ei.so,summationi=1ton of vi=2cardinality of E, summationi=1ton of ei=3f(G) ,summationi=1ton of ti=4g(G) note ei<=vi(vi-1)/2 and ei<=cardinality of E,so ei62,=vi^2*cardinality of E/2,i.e.ei<=vi*cardinality of E/261/2.summing over alli f(G)^2<=2(cardinality of E)^3/9 so, ti=f(Gi)<=(2/9)^1/3 f(G)61/38e summing all such inequalities we are done.and 3/32is best possible taking complete graph and limit. i

First a lemma A graph has $f$ triangles and $h$ edges, then $f^2 \le (2/9)h^3$. Proof: Let $d_i$ be degree of vertex $i$. Then that vertex is in at most $h$ triangles, but also at most $d_i^2/2$ triangles. Thus, $3f \le \sum \sqrt{hd_i^2/2} = \sqrt{h/2} \sum d_i = \sqrt{2h^3}$, which implies result. Now back to problem. Graph has $f$ triangles and $g$ K4s. Let vertex $i$ be in $f_i$ triangles and $g_i$ K4s. Then consider graph of neighbours of vertex $i$. Applying lemma to it, since it has $f_i$ edges and $g_i$ triangles, then $g_i^2 \le (2/9) f_i^3$. But also $g_i \le f$ trivially. So we get $ 4g = \sum g_i \le \sum ((2/9)f_i^3f)^{1/3} = (2f/9)^{1/3} * \sum f_i = (2f/9)^{1/3} * (3f)$ and so $ 64g^3 \le (2f/9) * (27f^3) = (2*27)f^4$ and so $g^3 \le (3/32)f^4$. Equality is for big cliques. Answer is $3/32$.

BS post, ignore please.

A Generalization from a book - let $N_t$ be the number of $K_t$s present in $G$. Prove that $\frac{N^t_{t+1}}{N^{t+1}_t} \le \frac{t!}{(t+1)^t}$. For $t=3$ we have this problem. @below Indeed, the book does not mention the equality at all - it just says that the proof of this inequality is similar to the ISL problem itself. @below again Nope, it's from a Korean combo book.

Then again the same argument proves that $c \ge \frac{t!}{(t+1)^t}$ but proving equality seems hard. @above Is this from "Straight from the Book"?

rkm0959 wrote: A Generalization from a book - let $N_t$ be the number of $K_t$'s present in $G$. Prove that $\frac{N^t_{t+1}}{N^{t+1}_t} \le \frac{t!}{(t+1)^t}$ Proof is by induction: Base Case: For $t = 1$, the conclusion is clear. We wish to show $|E| \le \frac{1} {2} n^2$, where $n$ is the number of $K_1$'s, i.e. vertices, in $G$ (assuming $G$ has $n$ vertices). $|E| \le {{n} \choose {2}} \le \frac{1}{2} n^2$, with equality occurring at $K_n$ for sufficiently large $n$, where $\frac {n^2 - n} {2}$ behaves like $\frac {n^2} {2}$. Inductive Step: Assume that $\frac{N^{t-1}_{t}}{N^{t}_{t-1}} \le \frac{(t-1)!}{t^{t-1}}$ holds for $t-1$. We will now show that the inequality holds for $t$. Indeed, consider the graph $G$ with vertices $v_1, v_2, ..., v_n$ and fix vertex $v_i$. Let $A_i$ be the number of $K_{t+1}$'s that $v_i$ belongs to. We can bound this number in two ways. On one hand, consider the subgraph induced by the neighbor set of $v_i$. $B_i$, the number of $K_{t}$'s contained in this subgraph, is also the number of $K_{t+1}$'s $v_i$ belongs to. Let $C_i$ be the number of $K_{t-1}$'s contained in the subgraph. By the inductive hypothesis, $ \frac{B^{t-1}_i}{C^t_i} \le \frac{(t-1)!}{t^{t-1}}$. It follows that $A^{t-1}_i \le \frac{(t-1)!}{t^{t-1}} \cdot C^t_i$ The second way of bounding $A_i$ is simpler. $A_i \le \frac {\sum_{v_j \perp v_i} B_j} {t} \le \frac {\sum_{v_i} B_i} {t} = \frac{t \cdot N_t} {t} = N_t$, since for each neighbor the number of $K_t$'s it is in is counted $t$ times, and the sum of $B_i$'s count $N_t$ $t$ times. Combining our bounds, $A^t_i \le \frac{(t-1)!}{t^{t-1}} \cdot C^t_i \cdot N_t$. Then $A_i \le C_i \cdot \sqrt [t] {\frac{(t-1)!}{t^{t-1}} \cdot N_t}$. Thus, $N_{t+1} = \frac {1} {t+1} \cdot {\sum_{v_i} A_i} \le \frac {1} {t+1} \cdot \sqrt [t] {\frac{(t-1)!}{t^{t-1}} \cdot N_t} \cdot {\sum_{v_i} C_i}$. But notice that ${\sum_{v_i} C_i} = t \cdot N_t$, since the number of $K_t$'s $v_i$ is part of is $C_i$, so summing over all vertices and multiplying by $t$ since each $K_t$ is counted $t$ times gives us this relation. Then $N_{t+1} \le \frac {1} {t+1} \cdot t \cdot N_t \cdot \sqrt [t] {\frac{(t-1)!}{t^{t-1}} \cdot N_t}$. Raising both sides to the $t$th power, we obtain: $N^t_{t+1} \le \frac{(t-1)!}{t^{t-1}} \cdot N^{t+1}_t \cdot \frac {t^t} {(t+1)^t} = \frac {t!} {(t+1)^t} \cdot N^{t+1}_t$, as desired. Conclusion: $N^t_{t+1} \le \frac {t!} {(t+1)^t} \cdot N^{t+1}_t$ for all $t$. To show this constant is optimal, simply take $K_N$ for $N$ sufficiently large, and note equality occurs.

orl wrote: For a finite graph $G$, let $f(G)$ be the number of triangles and $g(G)$ the number of tetrahedra formed by edges of $G$. Find the least constant $c$ such that \[g(G)^3\le c\cdot f(G)^4\]for every graph $G$. Proposed by Marcin Kuczma, Poland Let $V, E$ denote the vertex set and edge set of graph $G$. Let $\nu_i$ denote the $i^{th}$ vertex. Let $G_i$ be the subgraph induced by $\nu_i$ and let $V_i, E_i$ be it’s vertex set and edge set. Denote by $T_i$ to be the number of $3$-cycles in $G_i$ and let the total number of triangles and tetrahedrons in $G$ be $X, Y$ respectively. Claim : $X \leq \frac{2 \cdot |E|^{3/2}}{3 \sqrt{2}}$ Proof. Note that $|E_i| \leq \tbinom{|V_i|}{2} \leq \tfrac{|V_i|^{2}}{2}$. Also note that $\sum{|E_i|} = 3 \cdot X$. So, we have $$3 \cdot X = \sum{|E_i|} = \sum{\sqrt{|E_i|} \cdot \sqrt{|E_i|}} \leq |E|^{1/2} \cdot \sum{\tfrac{|V_i|}{\sqrt{2}}} = \tfrac{2 \cdot |E|^{3/2}}{\sqrt{2}}$$$\square$ By Claim in graph $G_i$, we obtain that $T_i \leq \tfrac{2 \cdot |E_i|^{3/2}}{\sqrt{2}}$. So, we have $T_i = T_i^{1/3} \cdot T_i^{2/3} \leq \tfrac{2^{1/3} \cdot |E_i| \cdot X^{1/3}}{3^{2/3}}$. Also note that we have $\sum{T_i} = 4Y$ and $\sum{|E_i|} = 3X$. So, we have $4Y = \sum{T_i} \leq \tfrac{3 \cdot 2^{1/3} \cdot X^{4/3}}{3^{2/3}}$ which in turn rearranges to $X^3 \leq \frac{3}{32} \cdot Y^4$. We claim that $\mathcal{C} = \tfrac{3}{32}$ is the smallest possible one. Indeed note that the value of $\tfrac{X^3}{Y^4}$ for $K_n$, the $n$-vertex complete graph, as $n \mapsto \infty$ “approaches” $\frac{3}{32}$. $\blacksquare$

yayups wrote: \begin{align*} g &\le \frac{\sqrt{2}}{12}\sum_{w\in V}|E_w|^{3/2} \\ &\textcolor{red}{\le \frac{\sqrt{2}}{12}|V|\left(\frac{3f}{|V|}\right)^{3/2}} \\ &\vdots\end{align*} Can you explain this step? I think the reverse inequality is true.

ayan.nmath wrote: yayups wrote: \begin{align*} g &\le \frac{\sqrt{2}}{12}\sum_{w\in V}|E_w|^{3/2} \\ &\textcolor{red}{\le \frac{\sqrt{2}}{12}|V|\left(\frac{3f}{|V|}\right)^{3/2}} \\ &\vdots\end{align*} Can you explain this step? I think the reverse inequality is true. oops yea, i think my solution is fake I'll fix it at some point.

The answer is \(\tfrac3{32}\), attained by arbitrarily large cliques. Let the vertices of \(G\) be \(v_1\), \(\ldots\), \(v_n\), and let \(e_i\) edges contain \(v_i\), let \(f_i\) triangles contain \(v_i\), and let \(g_i\) tetrahedrons contain \(v_i\). Also let \(G\) contain \(E\) edges, \(F\) triangles, and \(G\) tetrahedrons, so that \(2E_i=\sum e_i\), \(3F_i=\sum f_i\), \(4G_i=\sum g_i\). Claim: \(F^2\le\tfrac29E^3\). Proof. Note that \(f_i\le E\) and \(f_i\le\tbinom{e_i}2\). Then \[3F=\sum_{i=1}^nf_i\le\sum_{i=1}^n\sqrt{E\binom{e_i}2}<\sqrt{\frac E2}\sum_{i=1}^ne_i=\sqrt2E^{3/2},\]and the claim follows. \(\blacksquare\) Note the trivial bound \(g_i\le F\). Consider the subgraph of \(G\) comprised of the neighbors of \(v_i\) (but not \(v_i\)). It contains \(f_i\) edges and \(g_i\) triangles, so \(g_i^2\le\tfrac29f_i^3\). Hence \[4G=\sum_{i=1}^ng_i\le\sum_{i=1}^n\sqrt[3]{F\cdot\frac29f_i^3}=\sqrt[3]{\frac{2F}9}\sum_{i=1}^nf_i=3\sqrt[3]{\frac29}F^{4/3},\]and the desired bound follows. Remark: Repeating the argument presented above gives the following generalization: if \(A_t\) is the number of \(K_t\)'s formed by edges of \(G\), then \[\frac{A_{t+1}^t}{A_t^{t+1}}\le\frac{t!}{(t+1)^t}.\]

We claim the best bound is $\frac{3}{32}$, which can be easily attained by large cliques. First, define $e(G)$ to be the number of edges. We claim that $f(G)^2\le \frac{2}{9}e(G)^3$ We prove this with induction. Consider adding a vertex, $v$, to $G$ which already satisfies the condition and the induced subgraph $H$ which consists of vertices incident to an edge through $v$. The addition of $v$ adds $v(H)$ edges at at most $e(H)$ triangles, so we wish to show that $$(f(G)+e(H))^2\le \frac{2}{9}(e(G)+v(H))^3$$Of course, it suffices to assume $f(G)=\sqrt{\frac{2}{9}e(G)^3}$, and $v(H)=\sqrt{2e(H)}$ to stricten the inequality. Now, if we let $x=e(G)$, $y=e(H)$, we wish to show $\frac{2}{9}\left(x+\sqrt{2y}\right)^3\ge\left(\sqrt{\frac{2}{9}x^3}+y\right)^2\iff 3x^2\sqrt{2y}+6xy+2y\sqrt{2y}\ge 3xy\sqrt{2x}+\frac{9}{2}y^2$. However, as $x\ge y$, $3x^2\sqrt{2y}\ge 3xy\sqrt{2x}$ and $6xy>\frac{9}{2}y^2$, so we immediately get the desired result. Now, we proceed in the exact same fashion. Suppose that graph $G$ already satisfies the inequality, and consider adding a vertex $v$, incident upon an induced subgraph $H$. We add at most $e(H)$ new triangles and $f(H)$ tetrahedra, so we wish to show $$\left(\sqrt[3]{\frac{3}{32}x^4}+y\right)^3\le \frac{3}{32}\left(x+\sqrt[3]{\frac{9}{2}y^2}\right)^4$$where $x=f(G)$ and $y=f(H)$. Expanding once again, this becomes $$3y\sqrt[3]{\frac{9}{1024}x^8}+3y^2\sqrt[3]{\frac{3}{32}x^4}+y^3\le \frac{3}{8}x^3\sqrt[3]{\frac{9}{2}y^2}+\frac{9}{16}x^2\sqrt[3]{\frac{81}{4}y^4}+\frac{27}{16}xy^2+\frac{27}{64}y^2\sqrt[3]{\frac{9}{2}y^2}$$This is true because $\frac{3}{8}x^3\sqrt[3]{\frac{9}{2}y^2}\ge 3y\sqrt[3]{\frac{9}{1024}x^8}$, $\frac{9}{16}x^2\sqrt[3]{\frac{81}{4}y^4}>3y^2\sqrt[3]{\frac{3}{32}x^4}$, and $\sqrt{\frac{27}{16}}xy^2>y^3$

Use $A$ and $B$ in place of $f(G)$ and $g(G)$ respectively. Lemma: Show that in a graph with $m>0$ edges, there are fewer than $\frac{\sqrt{2}}{3}m^{3/2}$ triangles. Solution: The number of triangles is at most \[\frac{1}{3}\sum_v \min\left(\tbinom{\deg v}{2},m\right) < \frac{1}{3}\sum_v \deg v\sqrt{m/2} = \frac{\sqrt{2}}{3}m^{3/2}.\]Now, onto the problem itself. We claim $c=\tfrac{3}{32}$ does the job. Taking large cliques $K_n$, $B^3$ is asymptotically $\frac{n^{12}}{24^3}$ and $A^4$ is asymptotically $\frac{n^{12}}{6^4}$, so a lower constant than $\tfrac{3}{32}$ cannot work. Now it remains to show \[B^3\le \frac{3A^4}{32}.\]If $A>0$, we actually claim \[B<\frac{1}{4}\sqrt[3]{6}A^{4/3}.\]Let $f(v)$ for a vertex $v$ denote the total number of times the vertex is contained in a triangle, so $\displaystyle\sum_v f(v)=3A$. By the lemma, a vertex $v$ is part of at most \[\min\left(\frac{\sqrt{2}}{3}m^{3/2},A\right)\]tetrahedrons. Thus, we have \[B\le \frac{1}{4}\sum_v \min\left(\frac{\sqrt{2}}{3}m^{3/2},A\right) < \frac{1}{4}\sum_v \sqrt[3]{\frac{2}{9}}m\cdot A^{1/3} = 3A^{4/3}\cdot \frac{1}{4}\cdot \sqrt[3]{\frac 29} = \frac 14\sqrt[3]{6}A^{4/3},\]as desired.

I claim our answer is $c = \tfrac {3}{32}$, which can be achieved when we let $G$ be a complete graph $K_n$ and take $n \to \infty$. Indeed, note\[\frac{g(G)^3}{f(G)^4} \approx \frac{(n^2/24)^3}{(n^3/6)^4} = \frac{3}{32}\]for super large $n$. Let $G$ consist of $n$ vertices $\{v_i\}_{i = 1}^n$. Each vertex has degree $d_i$. First, note that $e(G)$, the number of edges, always must satisfy $f(G)^2 \leq \tfrac29e(G)^3$. Indeed, for every vertex, if it is in $t_i$ triangles, then\[f(G) = \frac13\sum t_i\]where clearly $t_i \leq e(G) - d_i$ and $\tbinom{d_i}{2}$. Hence, we may stupidly bound by $t_i \leq \sqrt{(e(G) - d_i)\tbinom{d_i}{2}} \leq d_i\sqrt{\tfrac{e(G)}{2}}$. It follows that\[f(G) = \frac13 \sum t_i \leq \frac 13\sqrt{\frac{e(G)}{2}}\sum d_i = \sqrt{\frac29 e(G)^3}\]since $\sum d_i = 2e(G)$ is known. Squaring proves our result, which holds for any graph $G$. We can more or less, iterate this to get a similar result for $f(G), g(G)$. In our original graph $G$, say vertex $v_i$ is part of $t_i$ triangles and $T_i$ copies of $K_4$. Then\[g(G) = \frac14\sum T_i\]For each vertex $v_i$, examine the graph $G_i = G \backslash v_i$. Note $t_i$ counts the number of edges in $G_i$, and $T_i$ the number of triangles. Hence it follows that $T_i^2 \leq \tfrac29 t_i^3$. Clearly $T_i = f(G\backslash v_i) \leq f(G)$ so note $T_i^3 \leq \tfrac29 t_i^3f(G)$. Thus,\[g(G) = \frac14\sum T_i \leq \frac14\sqrt[3]{\frac{2f(G)}{9}}\sum t_i = \sqrt[3]{\frac{3}{32}f(G)^4}\]which directly leads to our desired result. Remark: In fact, continuously iterating like this actually yields that $C_{n+1}(G)^n \leq \tfrac{n!}{(n+1)^n}C_n(G)^{n+1}$ where $C_i(G)$ counts the number of copies of $K_i$ in $G$.

Stronger C8 wrote: Given a graph $G$ and an integer $t\geqslant 3,$ let $K_t(G)$ denote the number of copies of $K_t$ in $G.$ Fix a positive integer $n$. Find the least constant $C_n$ such that for any graph $G$ we have \[K_{n+1}(G)^n\leqslant C_n\cdot K_n(G)^{n+1}.\] We will prove that $C_n=n!/(n+1)^n.$ To show that the inequality holds, we will proceed by induction. Let's firstly prove the base case, $n=2.$ Lemma. Let $G$ be a graph with $e$ edges and $t$ triangles. Then, $T^2\leqslant 2/9\cdot E^3.$ Proof. For every vertex $v_i,$ let $T_i$ denote the number of triangles containing $i$ and $E_i$ denote the number of edges in the neighbourhood of $v_i.$ Consequently, \[T_i=E_i=\sqrt{E_i\cdot E_i}\leqslant \sqrt{E\cdot \binom{\deg(v_i)}{2}}\leqslant\sqrt{\frac{E}{2}}\cdot \deg(v_i).\]By summing the latter over all vertices $v_i,$ it follows that\[3T=\sum_{i}T_i\leqslant\sqrt{\frac{E}{2}}\cdot \sum_{i}\deg(v_i)=\sqrt{\frac{E}{2}}\cdot 2E\]and thus, $T^2\leqslant 2/9\cdot E^3,$ as desired. Now, assume that the inequality holds for $n-1.$ For every vertex $v_i,$ let $K_t(v_i)$ denote the number of $K_t$'s which contain $v_i.$ Furthermore, let $N(v_i)$ denote the neighbourhood of $v_i.$ Then, \[K_{n+1}(v_i)=K_n(N(v_i))=\sqrt[n]{K_n(N(v_i))\cdot K_n(N(v_i))^{n-1}}\leqslant \sqrt[n]{K_n(G)\cdot C_{n-1}}\cdot K_{n-1}(N(v_i))\]As seen previously, by summing the latter over all vertices $v_i$ it follows that \begin{align*}(n+1)\cdot K_{n+1}(G) &=\sum_i K_{n+1}(v_i)\leqslant \sqrt[n]{K_n(G)\cdot C_{n-1}}\cdot \sum_iK_{n-1}(N(v_i)) \\ &=\sqrt[n]{K_n(G)\cdot C_{n-1}}\cdot n\cdot K_n(G)\end{align*}And, by substituting $C_{n-1}=(n-1)!/n^{n-1},$ we finally get that \[K_{n+1}(G)^n\leqslant \frac{n^n\cdot C_{n-1}}{(n+1)^n}\cdot K_n(G)^{n+1}=\frac{n!}{(n+1)^n}\cdot K_n(G)^{n+1}.\]Therefore, the inequality also holds for $n,$ completing the induction. Now, we shall prove that $C_n=n!/(n+1)^n$ is indeed optimal. To do so, notice that \[\lim_{t\to\infty}\frac{K_{n+1}(K_t)^n}{K_n(K_t)^{n+1}}=\lim_{t\to\infty}\binom{t}{n+1}^n\binom{t}{n}^{-(n+1)}=\frac{n!^{n+1}}{(n+1)!^n}=\frac{n!}{(n+1)^n}.\]

I will prove just the normal version. The generalization is basically the same proof. Notation: for any discrete random variable $X$, denote its entropy by $$H(X) := -\sum_{s} \mathbb P(X=s)\log\mathbb P(X=s).$$ The answer is $\boxed{3/32}$, achieved by taking an arbitrarily large clique. To prove the bound, take a uniform (ordered) $K_4$ subgraph $WXYZ$ of $G$. Then, using Shearer's Lemma, we get \begin{align*} 3\log(24g(G)) &= 3H(W,X,Y,Z) \\ &\leq H(W,X,Y) + H(X,Y,Z) + H(Y,Z,X) + H(Z,X,W) & \text{(Shearer)}\\ &= 4H(X,Y,Z) \\ &\leq 4\log(6f(G)), &\text{(uniform bound)} \end{align*}giving $(24g(G))^3 \leq (6f(G))^4$, proving the bound for $c=6^4/24^3 = 3/32$.

Wow, such an impressive application of Shearer's lemma - never would have thought of this!

Let $f_r(G)$ denote the number of $K_r$'s in a graph $G$. We prove a more general statement: for all graphs $G$, \[ f_r(G)\leq c_r f_{r-1}(G)^\frac{r}{r-1} \]where \[ c_r=\frac{(r-1)!^\frac{r}{r-1}}{r!}. \] We proceed by induction on $r$. For $r=2$, the statement is \[ ||G||\leq\frac{1}{2}|G|^2 \]which is true. Assume the statement for $r$. Let $G$ be an arbitrary graph. For every vertex $v$, let $t_r(v)$ denote the number of $K_r$'s containing $v$. By the induction hypothesis, \[ t_r(v)=f_{r-1}(G[N(v)])\leq c_{r-1}f_{r-2}(G[N(v)])^\frac{r-1}{r-2}=c_{r-1}t_{r-1}(v)^\frac{r-1}{r-2}. \]We also trivially have \[ t_r(v)=f_{r-1}(G[N(v)])\leq f_{r-1}(G). \]It follows that \[ t_r(v)\leq\min\left\{c_{r-1}t_{r-1}(v)^\frac{r-1}{r-2},f_{r-1}(G)\right\}\leq\sqrt[r-1]{c_{r-1}^{r-2}t_{r-1}(v)^{r-1}f_{r-1}(G)} \]so \begin{align*} f_r(G)&=\frac{1}{r}\sum_{v\in V}t_r(v)\\ &\leq\frac{1}{r}\sum_{v\in V}\sqrt[r-1]{c_{r-1}^{r-2}t_{r-1}(v)^{r-1}f_{r-1}(G)}\\ &=\frac{1}{r}c_{r-1}^{r-2}f_{r-1}(G)^\frac{1}{r-1}\sum_{v\in V}t_{r-1}(v)\\ &=\frac{r-1}{r}c_{r-1}^\frac{r-2}{r-1}f_{r-1}(G)^\frac{r}{r-1}\\ &=c_r f_{r-1}(G)^\frac{r}{r-1}. \end{align*}The induction step is complete. Note that \[ \lim_{n\rightarrow\infty}\frac{f_r(K_n)}{f_{r-1}(K_n)^\frac{r}{r-1}}=c_r \]so this constant cannot be improved.

The key lemma is the following: Lemma. In a graph with $m$ edges, there are at most $\frac{\sqrt 2}3 m^{3/2}$ triangles. Notice that at some vertex $v$, at most $\min\left\{{\deg v \choose 2}, m\right\}$ triangles emanate from $v$. Thus we have the bound $$T \leq \frac 13 \sum_{v \in G}\min\left\{{\deg v \choose 2}, m\right\} < \frac{\sqrt 2}3 m^{3/2}$$triangles, as needed. $\blacksquare$ Now pick some vertex $v$ and consider all the neighbors of $v$. There are at most $f(v) = \frac{\sqrt 2}3 (T(v))^{3/2}$ triangles among these neighbors, where $T(v)$ is the number of triangles at $v$; as a result, there are $\min(A, f(v))$ tetrahedrons at $v$. Then, $$4B \leq A^{\frac 13} \sum_{v \in G} \sqrt[3]{\frac 29} T(v) =\frac{\sqrt[3]6}4 A^{4/3},$$which proves the result.

The answer is $\frac{3}{32}$ which is achieved by $K_n$ as $n \rightarrow \infty$. Claim 1: The graph has at most $\frac{\sqrt{2}}{3}|E|^{3/2}$ triangles.

Now consider the neighbourhood of a vertex $\nu$, call it $H(\nu)$, then by claim there are at most $|K_4| \leq \sum \frac{\sqrt{2}}{3} {|H(\nu)|}^{3/2} \leq \sqrt[3]{6}/4 |T|^{4/3}$. Done.

We will solve the general version. Quote: Let $N_m$ be the number of $K_m$'s present in $G$. Prove that \[\frac{N^m_{m+1}}{N^{m+1}_m} \le \frac{m!}{(m+1)^m}\] Obviously we will induct on $m$. The base case $m=1$ is trivial. So assume this is true for $m=t-1$, we will prove this for $m=t$. Look at a vertex $v$ of $G$ and the induces subgraph by its neighbor set. Say it has $(M_v)_{t-1}$ number of $K_{t-1}$; this means by the inductive hypothesis it has atmost \[\frac{(t-1)!^{\frac t{t-1}}}t \cdot (M_v)_{t-1} \text{ number of } K_t\]There is anyways maximum $N_t$ number of $K_t$, $A$ makes and hence \begin{align*} (t+1)N_{t+1} & \leq \sum_v \min \left(\frac{(t-1)!^{\frac t{t-1}}}t \cdot (M_v)_{t-1},N_t \right) \\ & \leq \sum_v \left(\frac{(t-1)!^{\frac t{t-1}}}t \right)^\frac{t-1}t \cdot (M_v)_{t-1} \cdot N_t^\frac 1t \\ &=tN_t \cdot \frac{(t-1)!^\frac 1t}{t^\frac{t-1}t} \cdot N_t^\frac 1t \\ \implies \frac{N_{t+1}^t}{N_t^{t+1}} & \leq \frac{t!}{(t+1)^t} \end{align*}And we are done. Remark 1: One can actually notice that this inequality is (asymptotically) sharp, by taking arbitrarily large complete graphs. Remark 2 (Alternative Source): I think this problem is also in Straight from the Book.