Let $C$ denote the number of triangles in $K_n$ with two edges the same color and the third edge a different color. Let $r_i, g_i, b_i$ be the number of red, green, and blue edges (respectively) emanating from vertex $i$. Let $c_i$ and $a_i$ denote the number of $C$ triangles and $A$ triangles vertex $i$ is a part of (respectively). Notice that $$c_i+a_i=\binom{r_i}{2}+\binom{g_i}{2}+\binom{b_i}{2}$$Summing over $i$, $$\sum_{i}(c_i+a_i)=C+3A=\sum_{i}\left(\binom{r_i}{2}+\binom{g_i}{2}+\binom{b_i}{2}\right)$$Because each $C$ triangle is counted only by the vertex connecting the two edges with the same color, and each $A$ triangle is counted by each vertex. Now by Jensen, $$C+3A\ge 3n\binom{\frac{2\binom{n}{2}}{3n}}{2}=\binom{n}{3}-\frac{n(n-1)}{3}$$Because $\sum_i (r_i+b_i+g_i)$ is just twice the number of edges in $K_n$. Noting $A+B+C=\binom{n}{3}$ finishes the problem.

Very similar flavor as 2016 C3.

Number the vertices from $1$ to $n$ and let $r_i$ be the number of red edges that vertex $i$ has. Define $g_i$ and $b_i$ similarly. Also let $C$ be the number of triangles with exactly two distinct colors. By double-counting unordered pairs of adjacent edges, we have \begin{align*} 3A+C&=\sum_{i=1}^n \left(\binom{r_i}{2}+\binom{g_i}{2}+\binom{b_i}{2}\right)\\ 3B+2C&=\sum_{i=1}^n (r_ig_i+g_ib_i+b_ir_i).\\ \end{align*}It suffices to prove that $3B+2C\leq 6A+2C+n(n-1)$, or \begin{align*} \sum_{i=1}^n (r_ig_i+g_ib_i+b_ir_i)\leq 2\sum_{i=1}^n \left(\binom{r_i}{2}+\binom{g_i}{2}+\binom{b_i}{2}\right)+n(n-1) &=\sum_{i=1}^n (r_i^2+g_i^2+b_i^2)-\sum_{i=1}^n (r_i+g_i+b_i)+n(n-1). \end{align*}But we have $\sum_{i=1}^n (r_i^2+g_i^2+b_i^2) \geq \sum_{i=1}^n (r_ig_i+g_ib_i+b_ir_i)$ by Muirhead, and furthermore $\sum_{i=1}^n (r_i+g_i+b_i)=\sum_{i=1}^n \mathrm{deg}(i)=n(n-1)$, so we are done. $\blacksquare$ Remark: This problem is not extremely difficult, but I went down the wrong direction for a long time: I found a construction that almost achieved equality and thought it to be the best possible, so most of my work was using this "equality case" and could only have led nowhere. I was thinking about cherries, along with other things, the entire time, but in the wrong way. At some point, I noticed that for $n=4$ equality was achievable with something slightly different from the "equality case" but mostly believed that this was due to $n$ being small. However, I pushed this idea a bit further and constructed an equality case for $n=7$ that was basically done one edge at a time and had no discernible pattern whatsoever. At this point I realized that my previous specific ideas were all wrong and that the existence of "non-tractable" equality cases essentially meant that some global idea had to work (i.e. local optimizations wouldn't lead to anything, since the final state behaves poorly), and the problem died once I looked at cherries again.

Since the other two problems on this day can be solved using complex numbers, it is only fair to solve this problem using complex numbers as well. We'll also use some linear algebra. We begin with the following lemma: Lemma. Let $M_1$, $M_2$, and $M_3$ be symmetric matrices, and let $\omega=e^{2\pi i/3}$. Then \[\operatorname{tr}(M_1^3+M_2^3+M_3^3-3M_1M_2M_3)=\operatorname{tr}(M_1+M_2+M_3)(M_1+\omega M_2+\omega^2 M_3)(M_1+\omega^2 M_2+\omega M_3).\] Proof. The result holds if $M_1$, $M_2$, and $M_3$ commute, due to the factorization of the three-variable polynomial $x^3+y^3+z^3-3xyz$; in this case the matrices on the two sides would be equal. Expanding out both sides, this implies that the result would hold if the trace of a product were independent of the order of the terms in the product, i.e. if $\operatorname{tr}N_1N_2N_3=\operatorname{tr}N_{\sigma(1)}N_{\sigma(2)}N_{\sigma(3)}$ for any matrices $N_1$, $N_2$, $N_3$ and any permutation $\sigma$ of $\{1,2,3\}$. This is false in general, but it's true for symmetric matrices. The result is true for cyclic permutations since $\operatorname{tr}AB=\operatorname{tr}BA$, and it's true for a transposition: \[\operatorname{tr} N_1N_2N_3=\operatorname{tr}(N_1N_2N_3)^\intercal=\operatorname{tr} N_3^\intercal N_2^\intercal N_1^\intercal = \operatorname{tr} N_3N_2N_1.\]This finishes the proof. Now, we'll use this to solve the problem. Let $X$ be the $n\times n$ $\{0,1\}$-matrix defined so that $X_{ij}=1$ if and only if edge $ij$ is red (all diagonal entries of $X$ are zero), and define $Y$ and $Z$ similarly. Let $J$ be the $n\times n$ all-ones matrix, and let $I$ be the $n\times n$ identity matrix. We note $X$, $Y$, and $Z$ are symmetric, since $ij$ and $ji$ refer to the same edge; $X+Y+Z=J-I$, since each edge has exactly one color; $\operatorname{tr}(X^3+Y^3+Z^3)=6A$, since \[\operatorname{tr}(X^3)=\sum_{i,j,k=1}^n X_{ij}X_{jk}X_{ki}=\#\{(i,j,k)\colon ijk\text{ is a monochromatic red triangle}\}\]counts every red triangle six times; $\operatorname{tr}(XYZ)=B$, since \[\operatorname{tr}(XYZ)=\sum_{i,j,k=1}^n X_{ij}Y_{jk}Z_{kj}=\#\{(i,j,k)\colon \text{$ij$ red, $jk$ blue, $ki$ green}\}.\] So, letting $M=X+\omega Y+\omega^2Z$ and using our lemma, $$6A-3B=\operatorname{tr}(X^3+Y^3+Z^3-3XYZ)=\operatorname{tr}(J-I)MM^*=\operatorname{tr}JMM^*-\operatorname{tr}MM^*,$$where $M^*$ denotes the conjugate transpose of $M$. We need to show that this is at least $-n(n-1)$. Indeed, we have $$\operatorname{tr}MM^*=\sum_{i=1}^n\sum_{j=1}^n|M_{ij}|^2=n(n-1),$$since the diagonal entries of $M$ are zero and the off-diagonal entries of $M$ have magnitude $1$. So, we need $\operatorname{tr}JMM^*\geq 0$. Letting $\mathbf{1}$ denote the length-$n$ all-ones vector, we compute $$\operatorname{tr}JMM^*=\operatorname{tr}\mathbf 1\mathbf 1^* MM^*=\operatorname{tr}\mathbf 1^*MM^*\mathbf 1=\mathbf 1^*MM^*\mathbf 1=(M^*\mathbf 1)^*(M^*\mathbf 1)=\|M^*\mathbf 1\|^2\geq 0,$$as desired.

Same as many above. Consider the neighbourhood of vertex $v_i$, with $r_i,g_i,b_i$ red, green, blue edges connected to $v$ respectively. Let $C$ be the number of triangles with two colours. Let $S_i$ be the set of triangles connected to $v_i$. Thus, if $a_i,c_i$ are the number of $1$-colour and $2$-colour triangles in $S_i$ then: $$a_i+c_i\geq {r_i\choose 2}+{g_i\choose 2}+{b_i\choose 2}$$so summing over all vertices $v_i$, each $1$-colour triangle is counted 3 times while each $2$-colour is counted once: $$3A+C=\sum_{i=1}^n (a_i+c_i)\geq \sum_{i=1}^n \left({r_i\choose 2}+{g_i\choose 2}+{b_i\choose 2}\right)\geq 3n\cdot \frac{(\frac{n-1}{3})\left(\frac{n-1}{3}-1\right)}{2}$$by Jensen's inequality. Further evaluation shows that $$3A+C\geq3n\cdot \frac{(\frac{n-1}{3})\left(\frac{n-1}{3}-1\right)}{2}=\frac{n(n-1)(n-4)}{6}={n\choose 3}-\frac{n(n-1)}{3}=A+B+C-\frac{n(n-1)}{3} $$so $$B\leq 2A+\frac{n(n-1)}{3}$$as needed.

Define $C$ to be the number of triangles with two distinct colored edges. Define $r_k, g_k, b_k$ to be the number of red, green, and blue edges containing vertex $k$. Summing globally we receive $$3A+C = \sum_{k} {r_k \choose 2} + {b_k \choose 2} + {g_k \choose 2}$$because each we are essentially taking two vertices of the same color (as represented by the RHS) and this causes an $A$ triangle to be counted thrice and a $C$ triangle to be counted once due to how there is only one vertex in a $C$ triangle that is part of edges that are the same color. By the same logic we receive $$3B+2C = \sum_{k} r_kb_k + b_kg_k + r_kg_k$$From doubling the first equation and subtracting the second equation, we receive $$3B - 6A = \sum_{k} r_k^2+b_k^2+g_k^2-(r_k+b_k+g_k) - r_kb_k - b_kg_k-r_kg_k$$Notice that $\sum_{k} r_k+b_k+g_k = n(n-1)$, so we are essentially trying to prove that $$r_k^2+b_k^2+g_k^2 \ge r_kb_k + b_kg_k+r_kg_k$$which is equivalent to $$(r_k-b_k)^2+(b_k-g_k)^2+(r_k-g_k)^2 \ge 0$$which is trivially true.

Let $C$ be the number of triangles in $K_n$ with two edges of one color, and the third edge of a different color. Also let $r_i, g_i, b_i$ denote the number of red, green, and blue edges adjacent to vertex $i$, respectively. Notice that $A + B + C = \binom{n}{3}$ since it is the number of triangles in $K_n$. Now, we also have $\sum_{i=1}^n (r_i + b_i + g_i) = 2\binom{n}{2}$ since it counts each edge twice. Additionally, \[ 3A + C = \sum_{i=1}^n \left( \binom{r_i}{2} + \binom{g_i}{2} + \binom{b_i}{2} \right) \]because it counts each pair of two edges with the same color. By Jensen with $f(x) = \frac{x(x-1)}{2}$, we have \begin{align*} 3A + C \ge 3n f\left( \frac{2 \binom{n}{2} }{3n} \right) \\ = f\left( \frac{n-1}{3} \right)\cdot 3n = \frac{n(n-1)(n-4)}{6} \\ \end{align*} Hence \begin{align*} 2A + \frac{n(n-1)}{3} - B = 3A +C + \frac{n(n-1)}{3} - \binom{n}{3} \\ \ge \frac{n(n-1)(n-4)}{6} + \frac{2n(n-1)}{6} - \binom{n}{3} = 0,\\ \end{align*}as desired.

hardstuck on so many dcw-global problems wtfmonkeywtf

Lets begin with fixing the naming convention. We let $A$, $B$, and $C$ denote the number of triangles consisting of $1$, $2$, and $3$ colors. Let $X$ count the pair of edges $(v_iv_j, v_iv_k)$ such that they are both the same color. Let $Y$ count the pair of edges $(v_iv_j, v_iv_k)$ such that they are different colors. Then clearly, $X + Y = \binom{n}{2}$ $3A + B = X \iff 2B + 3C = Y$ We now proceed by looking at a specific vertex $v_i$. Define $X_{v_i}$ and $Y_{v_i}$ in the obvious manner. Then let $\alpha$, $\beta$, and $\gamma$ denote the number of red, blue, and green edges with one vertex at $v_i$. Clearly we have, \begin{align*} X_{v_i} &= \binom{\alpha}{2} + \binom{\beta}{2} + \binom{\gamma}{2} \\ Y_{v_i} &= \binom{n-1}{2} - X_{v_i} \end{align*}However by Jensen's we must have, \begin{align*} X_{v_i} &\geq 3 \cdot \binom{\frac{n-1}{3}}{2}\\ Y_{v_i} &\leq \binom{n-1}{2} - 3 \cdot \binom{\frac{n-1}{3}}{2} \end{align*}Summing over all $i$ we have, \begin{align*} X &\geq 3n \cdot \binom{\frac{n-1}{3}}{2}\\ Y &\leq n \cdot \left( \binom{n-1}{2} - 3 \cdot \binom{\frac{n-1}{3}}{2} \right) \end{align*}Now this is equivalent to, \begin{align*} 3A + B &\geq 3n \cdot \binom{\frac{n-1}{3}}{2}\\ 2B + 3C &\leq n \cdot \left( \binom{n-1}{2} - 3 \cdot \binom{\frac{n-1}{3}}{2} \right) \end{align*}Multiplying the first equation by 2 and adding the inequalities we find, \begin{align*} 3C \leq 6A + n \cdot \left( \binom{n-1}{2} - 3 \cdot \binom{\frac{n-1}{3}}{2} \right) - 6n \cdot \binom{\frac{n-1}{3}}{2} \end{align*}which rearranges to the desired, \begin{align*} C \le 2A + \frac{n(n-1)}{3}. \end{align*}

what is da point of da title?

Alex-131 wrote: what is da point of da title? apparently at mop one of the classes covered how to solve problems of this exact type (like the form of solution method you should follow to solve it)

Let $C$ be the number of triangles with $2$ edges of one color and the third vertex of a different color. We double count the number of pairs of edges sharing a vertex and having the same color. Counting the triangles, we get that this is $3A+C,$ and since $A+B+C=\binom{n}{3}$ we get that this is $2A-B+\binom{n}{3}.$ Next, consider one vertex. Suppose $x$ of the edges from that vertex are red, $y$ are blue, and $z$ are green. We have $x+y+z=n-1,$ and by QM-AM we can get $\frac{x(x-1)+y(y-1)+z(z-1)}{2} \ge \frac{(n-1)(n-4)}{6},$ so there are at least that many pairs of the edges we are counting that share that vertex. Summing this across all vertices, we get $\frac{n(n-1)(n-4)}{6},$ so $2A-B+\frac{n(n-1)(n-2)}{6} \ge \frac{n(n-1)(n-4)}{6},$ and simplifying we get our desired result.

Write down variables for everything: $A$, $B$ and $C$ are the number of all-same, all-different, and 2-same 1-different triangles. $r_i$, $b_i$, and $g_i$ are the number of red, blue, and green edges from the vertex $i$. Then the conditions just reduce to \[ A + B + C = \binom{n}{3} \]which is from double counting the number of triangles by nodes or our variables, \[ \sum_i (r_i + b_i + g_i) = n(n-1) \]which is from using the handshake lemma, \[ 3A + C = \sum_i \left( \binom{b_i}{2} + \binom{r_i}{2} + \binom{g_i}{2} \right) \]which is from double counting triangles of type $A$ and $C$ via the nodes. In particular, because we overcount each $A$ triangle three times for its three vertices, while $C$ type triangles are counted only at the vertex that has the two same color edges. Then, we can use Jensen on $f(x) = \binom{x}{2}$ to find that \[ 3A + C \geq 3n\binom{\frac{n(n-1)}{3n}}{2} = 3n\binom{\frac{n-1}{3}}{2} = \frac{3n(n-1)(n-4)}{3\cdot 3\cdot 2} = \frac{n(n-1)(n-4)}{6} = \binom{n}{3} - \frac{n(n-1)}{3} \]But this implies that \[ 3A + C \geq A + B + C - \frac{n(n-1)}{3} \implies 2A + \frac{n(n-1)}{3} \geq B \]as desired. $\blacksquare$

Pretty Easy Problem Let $B$ be the number of triangles with 2 sides of same color and 1 of different then clearly we have, $$\sum \binom{r_i}{2} + \binom{g_i}{2} + \binom{b_i}{2} = 3A + B$$or $$\sum r_i^2+g_i^2+b_i^2 = 6A + 2B+n(n-1)$$and $$ \sum r_ig_i + g_ib_i + r_ib_i = 3C + 2B $$by $\sum r_i^2 + g_i^2 + b_i^2 \geq \sum r_ig_i+r_ib_i+g_ib_i$ (true by SOS) we are done.

Solved with Elliot_7002 Call the structure formed by two edges of same color sharing a common vertex to be monochromatic angle. Any monochromatic triangle (having all edges of same color) has $3$ monochromatic angles , any dichromatic triangle (whose edges are colored using only $2$ colors) has $1$ monochromatic angle , any trichromatic triangle (all edges of different color) has $0$ monochromatic angle. Let $N$ be the number of monochromatic angles. We have $N=3A+ \left(\binom{n}{3}-A-B\right)=\binom{n}{3}+2A-B$. It suffices to prove that $N \geq \binom{n}{3}-\frac{n(n-1)}{3}$. Consider any vertex $v_i$ of the graph having $a_i,b_i,c_i$ red,blue ,green edges incident to it. We have $a_i+b_i+c_i=n-1$. Number of monochromatic angles having vertex $v_i$ equals $\binom{a_i}{2}+\binom{b_i}{2}+\binom{c_i}{2}$. By Jensen inequality $\binom{a_i}{2}+\binom{b_i}{2}+\binom{c_i}{2} \geq 3 \cdot \binom{\frac{a_i+b_i+c_i}{3}}{2}= 3 \cdot \binom{\frac{n-1}{3}}{2}=\frac{(n-1)(n-4)}{6}$. We have $N=\sum_i \binom{a_i}{2}+\binom{b_i}{2}+\binom{c_i}{2} \geq \frac{n(n-1)(n-4)}{6}=\binom{n}{3}-\frac{n(n-1)}{3}$. So we are done

Since the two quantities aren't directly related to each other, the idea is to make a ``bridge" between them using some new quantity. So, in view of this, let $C$ be the number of triangles in $K_n$ with two edges of the same color, and one different. Also, let $a_i$, for any vertex $i$ be the number of edges connected to $i$ which are red, $b_i$ be the number of such edges which are blue, and $c_i$ be the number of such edges which are green. Note that we have $$\begin{aligned} A + B + C = \dbinom{n}{3} \\ 3A + C = \sum_{i = 1}^{n} \left( \dbinom{a_i}{2} + \dbinom{b_i}{2} + \dbinom{c_i}{2} \right)\end{aligned},$$then subtracting the two equations gives $$B - 2A = \dbinom{n}{3} - \sum_{i = 1} \left(\dbinom{a_i}{2} + \dbinom{b_i}{2} + \dbinom{c_i}{2}\right) \le \frac{n(n - 1)}{3},$$is the inequality we need, or $$\frac{n(n - 1)(n - 4)}{6} \le \sum_{i = 1} \left(\dbinom{a_i}{2} + \dbinom{b_i}{2} + \dbinom{c_i}{2}\right),$$but we have $a_i + b_i + c_i = n - 1,$ and $$\dbinom{a_i}{2} + \dbinom{b_i}{2} + \dbinom{c_i}{2} = \frac{a_i^2 - a_i + b_i^2 - b_i + c_i^2 - c_i}{2} = \frac{a_i^2 + b_i^2 + c_i^2 - (n - 1)}{2} \le \frac{3\left(\frac{a_i + b_i + c_i}{3}\right)^2 - (n - 1)}{2} = \frac{(n - 1)(n - 4)}{6},$$summing this up over all $i$ gives the result.

Surprisingly easy. Not posting since it is same as @above

Let $C$ denote the number of triangles with $2$ edges of $1$ color and the other edge with another color. Then, $A+B+C=\binom{n}{3}=\frac{n(n-1)(n-2)}{6}$. Let $r_i$, $g_i$, and $b_i$ represent the number of red, green, and blue edges coming out of vertex $i$, respectively. Then, $3A+C=\sum_{i=1}^n(\binom{r_i}2+\binom{g_i}{2}+\binom{b_i}2)\ge 3n\binom{\frac{n-1}{3}}{2}=\frac{n(n-1)(n-4)}{6}=\frac{n(n-1)(n-2)}{6}-\frac{n(n-1)}3$. Subtracting this from $A+B+C$ gives us $B-2A\le \frac{n(n-1)}3$, as desired.

For any vertex $i$ (call it $v_i$), let $r_i,b_i,g_i$ denote the number of red, blue and green edges which vertex $i$ is a part of. Let $C_i$ denote the number of triangles $v_i$ is in (say $v_iv_jv_k$), for which $v_iv_j$ and $v_iv_k$ are same color but $v_jv_k$ is different. Let $A_i$ denote the number of triangles $v_i$ is in such that all three edges are same color. Then, if $C$ denote the number of triangles with two edges one color and third edge as different color, then $C=\sum_{i}C_i$ and $3A=\sum_{i}A_i$ because $C$ is counted exactly once but $A$ is counted thrice, one from each vertex. Moreover, $C_i+A_i$ represents all triangles such that $v_iv_j$ and $v_iv_k$ are same color for pairs $(j,k)$. This is same as choosing two edges of $v_i$ of same color. Hence, \[C_i+A_i=\binom{r_i}2+\binom{b_i}2+\binom{g_i}2\]But we know, $r_i+b_i+g_i=n-1\Longrightarrow \frac{r_i^2+b_i^2+g_i^2-r_i-b_i-g_i}{2}\geq\frac{(n-1)(n-4)}{6}$ by applying Cauchy-Schwartz inequality. Summing over all such $n$ we get $C+3A\geq\frac{n(n-1)(n-4)}{6}$. Plugging $C=\frac{n(n-1)(n-2)}{6}-A-B$, we get the result. Remark. In general for any graph, a similar method gives \[3A+C\geq\frac{2e^2}{3n}-\frac e4\]

im xonin

For each vertex $i$, let $r_i$, $b_i$, and $g_i$ be the number of red, blue, and green edges. Also let $C$ be the triangles not counted in $A$ or $B$ (aka the ones with $2$ of the same colored edges) Counting the $A$ and $C$ triangles we have \[3A+C = \sum_{i = 1}^n {r_i \choose 2} + {b_i \choose 2}+{g_i \choose 2}\]We also know that \[\sum_{i = 1}^n r_i+b_i+g_i = 2 \cdot {n \choose 2}\]Now using Jensen's on $f(x) = \frac{x(x-1)}{2}$, we get \[3A+C \ge 3n \cdot f\left(\frac{2 \cdot {n \choose 2}}{3n}\right) = 3n \cdot \frac{n-1}{6} \cdot \frac{n-4}{3} = \frac{n(n-1)(n-4)}{6}\]Now using $A+B+C = {n \choose 3}$, we can simplify to the desired inequality.

me cooking, Let $C$ denote the number of triangles which have only 2 edges of common color. and $a_i,b_i,c_i$ be the number of red green blue vertices at the i'th vertex $$A + B + C = \binom{n}{3}$$Now we double count the pair of edges having same color so we get $$ 3A + C \geq \sum_{i=1}^n \left(\binom{a_i}{2}+ \binom{b_i}{2} + \binom{c_i}{2}\right)$$the function $f(x) = \frac{x(x-1)}{2}$ is convex so by jenson's ineq we have $$3A + C \geq \sum_{i=1}^n \left(\binom{a_i}{2}+ \binom{b _i}{2} + \binom{c_i}{2}\right) \geq \frac{n(\frac{n(n-1)}{3}) (\frac{n(n-1)}{3} - 1)}{2} = \frac{n(n-1)(n-4)}{6}$$So subtracting the first one from third we get $$ 2A - B \ge -\frac{n(n - 1)}{3}$$and hence \[ B\le 2A+\frac{n(n-1)}{3}.\]As required $\blacksquare$

We are going to define some new quantities. $\bullet$ Let $Z$ denote the number of triangles which have exactly $2$ side lengths same color and the last one different. $\bullet$ For the $i^\text{th}$ vertex, define $r_i$, $b_i$, $g_i$ to be the number of red, blue, and green edges forming from it respectively. $\bullet$ Let $X_i$ be the total number of monochromatic triangles formed with the $i^\text{th}$ vertex and $Z_i$ be the total number of triangles with exactly $2$ side lengths the same color and last different and the $2$ same colored side lengths meet at $i$. It is easy to check that we have the following equalities \begin{align*} &X+Y+Z=\binom{n}3\\ &r_i+b_i+g_i=n-1 \\ &\sum_i X_i=3X \text{ and } \sum_i Z_i=Z \end{align*}And finally easy to check we also have \begin{align*} & \qquad \text{ } \text{ } X_i+Z_i=\binom{r_i}2+\binom{b_i}2+\binom{g_i}2\\ &\implies 3X+Z =\sum_i \binom{r_i}2+\binom{b_i}2+\binom{g_i}2 \\ & \implies 3X+Z \overset{\text{JNS}}{\geq} 3n \cdot \frac{n-1}3 \cdot \frac{n-4}3 \cdot \frac12=\frac{n(n-1)(n-4)}6\\ &\implies 3X+Z-X-Y-Z \geq \frac{n(n-1)(n-4)}6-\binom{n}3=-\frac{n(n-1)}3\\ &\iff 2X+\frac{n(n-1)}3 \ge Y \end{align*}As desired.

Cool. Let $C$ denote the number of triangles with $2$ sides of the same color and the third side of a different color. Also call a vertex an apex of a triangle if its $2$ incident edges are of the same color (there can be multiple apexes in a triangle). Now look at a random vertex $u$. Say that $u$ has $r$ incident edges of color red and define $g$ and $b$ similarly. Then, the quantity $\binom{r}{2}+\binom{g}{2}+\binom{b}{2}$ counts all the triangles that have apex $u$. By Jensen, the quantity mentioned before is at least $3\binom{(n-1)/3}{2}$. Summing this over all vertices $u$ and noting that in this process every $C$ triangle is counted once and every $A$ triangle is counted $3$ times (since they have $3$ apexes), we get \[3A+C\ge 3n\binom{(n-1)/3}{2}=\frac{n(n-1)(n-4)}{6}\]Finally, writing $A+C=\binom{n}{3}-B$ yields the desired conclusion. $\blacksquare$

Similar to 2016 C3: Let $C$ denote no of triangles which have $2$ only two sides of same colour. Then: $$3A+C = \sum \binom{r_i}{2}+\binom{b_i}{2}+\binom{g_i}{2}$$$$3B+2C = \sum r_i b_i + b_i g_i + g_i r_i$$Thus, it suffice to show that: $$\sum r_i b_i + b_i g_i + g_i r_i \le \sum r_i (r_i-1)+ b_i (b_i-1) + g_i (g_i-1) + n(n-1).$$Note that: $$\sum r_i^2+b_i^2+g_i^2 \ge \sum r_i b_i + b_i g_i + g_i r_i$$and $\sum r_i+g_i+b_i = n(n-1)$ which follows the conclusion.

Let $C$ denote the number of the triangles with two of its edges having the same colour ,also in the $i$’th vertex, let $r_i,b_i$ and $g_i$ denote the number of edges of each colour incident to it. Then we have that. \[3A+C= \sum_{i=1}^{n} \binom{r_i}{2}+\binom{b_i}{2}+\binom{g_i}{2}\]In which by Jensen and the fact that $r_i+b_i+g_i=n-1$. We get \[ \sum_{i=1}^{n} \binom{r_i}{2}+\binom{b_i}{2}+\binom{g_i}{2} \geq 3n\binom{\frac{n-1}{3}}{2}\]Now as $A+B+C= \binom{n}{3}$ then substituting that in, we get the desired bound.

Extremely strong ISL 2016/C3 vibes. Let $K_n = (V, E)$ and for any $v \in V$, let $d_r(v)$ be the number of red edges that has $v$. Define $d_b(v), d_g(v)$ analogously. For any $u, v \in V$, let $d(u, v)$ be the number of vertices $w$ for which triangle $(u, v, w)$ has edges with same color. Then, obviously we have $\sum_{(u, v) \in E} d(u, v) = 3A$ and $\sum_{(u, v) \in E} ((d_{(u, v)-color}(u) - 1) + (d_{(u, v)-color}(v) - 1) - 2d(u, v)) = 2C$, where $C = \binom{n}{3} - A - B$ is the number of triangles that has exactly 2 same sides. $\implies \sum_{(u, v) \in E} (d_{(u, v)-color}(u) + d_{(u, v)-color}(v) - 2) = 6A + 2C = 4A + 2\binom{n}{3} - 2B \implies$ If we let $K = \sum_{(u, v) \in E} (d_{(u, v)-color}(u) + d_{(u, v)-color}(v) - 2)$, we get that $2B = 4A + 2\binom{n}{3} - K$. Combining with our inequality, we need to prove $2\binom{n}{3} - K \le \frac{2(n-1)n}{3} \Leftrightarrow K \ge 2(\binom{n}{3} - \frac{n(n-1)}{3})$. Now we compute $K$ as follows: $K = \sum_{(u, v) \in E} (d_{(u, v)-color}(u) + d_{(u, v)-color}(v) - 2) = -n(n-1) + \sum_{v \in V} (d_r(v)^2 + d_b(v)^2 + d_g(v)^2) \ge -n(n-1) + n \cdot \frac{(n-1)^2}{3} = 2(\binom{n}{3} - \frac{n(n-1)}{3})$, so we're done. $\blacksquare$ ,