gpt pro 5.2 attempt on Math-5742

1) FORMAL RESTATEMENT

For $k\ge 3$, the problem asks whether the following inequality holds:

For every graph $H$ with $m=e(H)$ edges and with no isolated vertices, \[ R(C_k,H) \le 2m+\left\lceil\frac{k-1}{2}\right\rceil. \]

2) QUICK LITERATURE/CONTEXT CHECK

The problem file notes that this statement was proved for even $k$ by Erdős, Faudree, Rousseau, and Schelp [EFRS93], and for $k=3$ by Sidorenko [Si93].

3) ATTACK PLAN

Before attempting a proof, check for small-$m$ obstructions. In particular, test $m=1$ (so $H=K_2$) because $R(C_k,K_2)$ is elementary.

4) WORK

Fast reality check.

Take $m=1$, so $H=K_2$. Then $R(C_k,K_2)=k$ (Lemma 570.1 below). For $k\ge 5$ the proposed upper bound evaluates to \[ 2\cdot 1+\left\lceil\frac{k-1}{2}\right\rceil = 2+\left\lceil\frac{k-1}{2}\right\rceil < k, \] so the inequality fails.

Lemma 570.1 (exact value for $H=K_2$). For every $k\ge 3$, \[ R(C_k,K_2)=k. \]

Proof. This is exactly Lemma 569.1 specialised to $\ell=k$. (The proof is repeated briefly.)

- In $K_{k-1}$ colour all edges red. Then there is no blue edge and no red $C_k$ (not enough vertices), so $R(C_k,K_2)>k-1$.

- In $K_k$, any colouring either has a blue edge (giving blue $K_2$) or is all red, in which case the red $K_k$ contains a $C_k$.

Hence $R(C_k,K_2)=k$. $\square$

Lemma 570.2 (the proposed inequality fails for all $k\ge 5$). For every $k\ge 5$, \[ R(C_k,K_2)=k > 2+\left\lceil\frac{k-1}{2}\right\rceil. \]

Proof. By Lemma 570.1, $R(C_k,K_2)=k$.

For $k\ge 5$ we have $\lceil (k-1)/2\rceil \le k/2$ (since $\lceil x\rceil\le x+1/2$ for half-integers, and here $(k-1)/2$ is either an integer or a half-integer). Thus \[ 2+\left\lceil\frac{k-1}{2}\right\rceil \le 2+\frac{k}{2}. \] Now $k>2+k/2$ is equivalent to $k/2>2$, i.e. $k>4$, which holds for all $k\ge 5$. Therefore \[ 2+\left\lceil\frac{k-1}{2}\right\rceil < k = R(C_k,K_2). \] So the claimed inequality fails when $H=K_2$ (which has $m=1$ and no isolated vertices). $\square$

Additional computed checks for $k=5$ and matchings. Although the conjectured inequality is false as stated, for $k=5$ and $m\ge 2$ it matches small computed values for matchings: \[ R(C_5,2K_2)=6 = 2\cdot 2+2,\qquad R(C_5,3K_2)=8=2\cdot 3+2. \]

5) VERIFICATION

- The counterexample uses $H=K_2$ (one edge, no isolated vertices) and the exact evaluation $R(C_k,K_2)=k$.

- The strict inequality $2+\lceil (k-1)/2\rceil<k$ is checked explicitly in Lemma 570.2.

- The numerical checks $R(C_5,2K_2)=6$ and $R(C_5,3K_2)=8$ were computed locally by exhaustive search/backtracking.

6) FINAL

LABEL: FULL SOLUTION

SUBLABEL: COUNTEREXAMPLE/DISPROOF

The statement "for all $k\ge 3$ and all $m$-edge graphs $H$ without isolated vertices, $R(C_k,H)\le 2m+\lceil (k-1)/2\rceil$" is false.

A concrete counterexample is: take any $k\ge 5$ and take $H=K_2$ (so $m=1$). Then $R(C_k,K_2)=k$ (Lemma 570.1), but the proposed bound gives $2+\lceil (k-1)/2\rceil<k$ (Lemma 570.2), contradicting the inequality.

(If one wants a corrected version consistent with standard conventions, a minimal repair is to add a hypothesis "$m$ is sufficiently large (depending on $k$)", or to replace the right-hand side by $\max\{k,\,2m+\lceil (k-1)/2\rceil\}$. The corrected statement is not proved here.)