#1 Basic Inequality Principles

We have \[2(a^2+b^2+c^2-ab-bc-ca)=(a-b)^2+(b-c)^2+(c-a)^2\ge0.\] Hence the inequality is true. Equality occurs exactly when \(a=b=c\).

\[x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}=98+\frac{(x-y)^2}{2}\ge98.\] The minimum is \(98\), attained at \(x=y=7\).

Since \(t^2+t+1>0\), we may multiply by \(3(t^2+t+1)\). We get \(3t\le t^2+t+1\), i.e. \((t-1)^2\ge0\). Equality occurs at \(t=1\).

Write the elements as \(a_1

We have \(up+vq-uq-vp=u(p-q)-v(p-q)=(u-v)(p-q)\). Both factors are positive, so the difference is positive.

Cyclically rename the variables so that \(x\ge z\). Then \(y+z\le y+x\), hence \(\frac{y}{y+z}\ge\frac{y}{y+x}\). Therefore \[\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge\frac{x}{x+y}+\frac{y}{y+x}+\frac{z}{z+x}=1+\frac{z}{z+x}>1.\]

Adding gives \(-x-y>0\), so \(1-x-y>1>0\). Now multiply the inequalities \(x^2-x>y^2\) and \(y^2-y>x^2\). This is legitimate because \(x^2\) and \(y^2\) are positive, and the left sides are larger than them. We obtain \((x^2-x)(y^2-y)>x^2y^2\). Expanding and cancelling gives \(xy(1-x-y)>0\). Since \(1-x-y>0\), we get \(xy>0\).

Since \(a+b+c=0\), \[0=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca).\] Therefore \[ab+bc+ca=-\frac{a^2+b^2+c^2}{2}\le0.\] Equality is possible only when \(a=b=c=0\).

Assume the contrary. Then for each \(i\), the discriminant \(4(p_i^2-q_i)\le0\), so \(p_i^2\le q_i\). Hence all \(q_i\ge0\), and the inequalities may be multiplied: \((p_1p_2p_3)^2\le q_1q_2q_3\). Thus \(N^2\le N\), impossible for \(N>1\). Therefore for some \(i\), \(p_i^2>q_i\), so the corresponding trinomial has two distinct real roots.

Let \(a_1\le a_2\le\cdots\le a_{2027}\). Since \(a_1+a_2+a_3\) is written on the board, it is at least \(a_1\), hence \(a_2+a_3\ge0\). Similarly, \(a_{2025}+a_{2026}+a_{2027}\) is written and is at most \(a_{2027}\), hence \(a_{2025}+a_{2026}\le0\). But \(a_2\le a_3\le a_{2025}\le a_{2026}\), so \[0\le a_2+a_3\le a_{2025}+a_{2026}\le0.\] Thus \(a_2+a_3=a_{2025}+a_{2026}=0\), and the ordering gives \(a_2=a_3=\cdots=a_{2026}=0\). Hence there are at least \(2025\) zeros. The example \(-1,0,\ldots,0,1\) shows the bound is sharp.

From \(a^3-b^3=3\) we get \(a>b\). Then \[3(a^2+b^2)=(a^2+b^2)(a^3-b^3)=a^5-b^5+a^2b^2(a-b).\] The right side is at least \(9\), because \(a^5-b^5\ge9\) and \(a^2b^2(a-b)\ge0\). Hence \(3(a^2+b^2)\ge9\), so \(a^2+b^2\ge3\).

Let \(a_1>a_2>\cdots>a_{12}>0\). Then \(a_1a_2a_3>a_4a_5a_6\). If \(a_6\ge1\), then \(a_1a_2a_3>a_4a_5a_6\ge a_4a_5\), so three numbers beat two. If \(a_6<1\), then \(a_7<1\), and \(a_1a_2a_3>a_4a_5a_6>a_4a_5a_6a_7\), so three numbers beat four.

Let the numbers be \(a_1,\ldots,a_{2n}\), with indices modulo \(2n\), and let \(S_i=a_i+a_{i+1}+\cdots+a_{i+n-1}\). The total sum is \(S>0\), so \(S_i+S_{i+n}=S\). Hence some \(S_i\) is positive. If all \(S_i>0\), any number works. Otherwise, while moving around the circle, there is a transition \(S_i>0\), \(S_{i+1}\le0\). Then the length-\(n\) arc ending at \(a_i\) from the other side has sum \(S-S_{i+1}>0\). Thus \(a_i\) is an endpoint of two arcs with positive sums.

Let \(m=37!\). Then \(38!=38m\), and \((38!)!=(38m)!\). Split \(1\cdot2\cdots38m\) into \(38\) blocks of length \(m\): in the \(j\)-th block, every factor is at most \(jm\), and at least one factor is smaller than \(jm\). Hence the product of the \(j\)-th block is strictly less than \((jm)^m\). Multiplying for \(j=1,\ldots,38\), \[(38m)!

If \(c\le950\), change \(c\) to \(0\). Then \(ax^2+bx=x(ax+b)\) has the integer root \(0\), and the cost is at most \(950\).

Now suppose \(c>950\). Then \(a+b<850\). Choose a square \(k^2\) nearest to \(c\). Since \(c<1800<43^2\), we may take \(k\le42\), and the distance from \(c\) to the nearest square is at most half the gap between neighbouring squares, hence at most \(42\). Change \(a\) to \(-1\), \(b\) to \(0\), and \(c\) to \(k^2\). The cost is at most \((a+1)+b+42<893\). The resulting trinomial \(-x^2+k^2\) has integer roots \(\pm k\). In both cases, \(950\) coins are enough.

Let \(M\) be the largest number. It cannot equal the absolute difference of its neighbours, because both neighbours are positive and at most \(M\). Hence \(M\) is the unique exception.

Choose a minimum \(d\). On one arc from \(d\) to \(M\), write \(d=a_0,a_1,\ldots,a_k=M\). By induction, \(a_{i+1}\ge a_i\), and since \(a_i=|a_{i+1}-a_{i-1}|\), in fact \(a_{i+1}=a_i+a_{i-1}\) for \(1\le i

The number \(d\) is not the exception, so \(d=|a_1-b_1|\). Suppose \(b_1>a_1\). Then \(b_1=a_1+d=a_2\). Continuing both recurrences gives \(a_0=b_0\le a_1

We need to prove \(x^9\ge16y^3\). From the condition, \(y^3\le x^5-4x\), so it is enough to show \(x^9\ge16x^5-64x\). But \[x^9-16x^5+64x=x(x^4-8)^2\ge0,\] since \(x>0\). Therefore \(x^9\ge16y^3\), hence \(x^3\ge\sqrt[3]{16}\,y\).

Assume \(A\cap B=\varnothing\). Then \(A\cup B\) contains \(2n\) distinct positive integers, so the sum of all its elements is at least \(1+2+\cdots+2n=n(2n+1)\). On the other hand, this sum equals \(n^2+n^2=2n^2\). But \(n(2n+1)>2n^2\), a contradiction. Hence a common element exists.

Adding gives \(x+y<0\). Suppose \(xy<0\). Without loss of generality, \(x>0\), \(y<0\). Then \(x+y<0\) implies \(00\). Hence \(xy\) is not negative; since \(x,y\ne0\), we get \(xy>0\).

Let the roots be \(r,s\), and suppose \(|r|\le1\), \(|s|>1\), since exactly one root lies in \([-1,1]\). By Vieta, \(rs=b>0\), so the roots have the same sign and \(b=|r||s|\). Since \(|s|>1\), we get \(|r|b\). Thus \(s\notin(-b,b)\). Therefore exactly one root lies in \((-b,b)\).

Call \(i\ge2\) bad if \(a_{i-1}a_{i+1}\) or \(a_{i-1}>a_i

Assume there are infinitely many bad indices, and choose one bad index \(k\). Take any \(n>k\). By the condition, \(a_n\) belongs to a monotone segment of length \(n+1\). Such a segment cannot contain all three terms \(a_{k-1},a_k,a_{k+1}\), because the direction changes there. Since the segment is so long and contains \(a_n\), it cannot start early enough to include \(a_{k-1}\); hence it must contain \(a_n,a_{n+1},a_{n+2}\). Thus \(n+1\) is not bad. Therefore for every \(n>k\), the index \(n+1\) is not bad, contradicting infinitely many bad indices. Hence there are only finitely many bad indices, and the tail is monotone.

For \(n=150,k=75\), we get \(a_{150}-a_{75}\ge150^3-75^3\), so \(a_{150}\ge150^3-75^3\). For \(n=75,k=150\), we get \(a_{75}-a_{150}\ge75^3-150^3\), so \(a_{150}\le150^3-75^3\). Therefore \[a_{150}=150^3-75^3=7\cdot75^3.\]

For every \(t\), \(t^2+t+1=\left(t+\frac12\right)^2+\frac34\ge\frac34\). Hence \(\frac1{t^2+t+1}\le\frac43\), with equality only at \(t=-\frac12\). The sum of three such fractions is at most \(4\). If it equals \(4\), each fraction must equal \(\frac43\), so \(x=y=z=-\frac12\), impossible for distinct numbers. Answer: no.

The number \(p_1=x_1-\frac1{x_1}\) is not a positive integer: if \(x_1=1\), it is \(0\), and if \(x_1>1\), it is not an integer. Hence at most \(11\) of the \(p_i\) can be positive integers.

We show that \(11\) is attainable. Take \(x_i=i+1\). Then \[x_i-\frac1{x_i}=i+1-\frac1{i+1}=\frac{i(i+2)}{i+1}.\] Therefore \[p_i=\frac{1\cdot3}{2}\cdot\frac{2\cdot4}{3}\cdots\frac{i(i+2)}{i+1}=\frac{i!(i+2)}2.\] For \(i\ge2\), this is a positive integer. Thus \(11\) of \(p_1,\ldots,p_{12}\) can be positive integers, and more is impossible.

From \(a^2+b^2<(a-b)^2=a^2-2ab+b^2\), we get \(ab<0\). Similarly, \(bc<0\). Thus \(a\) and \(b\) have opposite signs, and \(b\) and \(c\) have opposite signs; hence \(a\) and \(c\) have the same sign, so \(ac>0\). Then \[(a+c)^4-a^4-c^4=4a^3c+6a^2c^2+4ac^3=2ac(2a^2+3ac+2c^2)>0,\] because \(ac>0\) and \(2a^2+3ac+2c^2>0\). Therefore \(a^4+c^4<(a+c)^4\).

Answer: \(k(k-1)\). Construction: take one isolated vertex, split the remaining vertices into sets \(X\) and \(Y\) of sizes \(k\) and \(k-1\), and connect every vertex of \(X\) to every vertex of \(Y\). Then vertices in \(X\) have degree \(k-1\), vertices in \(Y\) have degree \(k\), the condition holds, and there are \(k(k-1)\) edges.

Now prove the upper bound. Let \(X_i\) be the set of vertices of degree \(i\), and let \(m\) be the maximum degree. If \(m\le k-1\), then \(2E\le2k(k-1)\), so \(E\le k(k-1)\).

If \(m\ge k+1\), choose a vertex of degree \(m\). All its neighbours have degree \(m-1\), so \(|X_{m-1}|\ge m\). Any vertex in \(X_{m-1}\) has \(m-1\) neighbours in \(X_m\cup X_{m-2}\), so the classes \(X_{m-1},X_m,X_{m-2}\) contain at least \(m+(m-1)>2k\) vertices, impossible.

It remains to consider \(m=k\). Edges connect only adjacent degree classes, so the graph is bipartite with parts \(Y=X_k\cup X_{k-2}\cup\cdots\) and \(Z=X_{k-1}\cup X_{k-3}\cup\cdots\). On one hand, \(E\le k|Y|\); on the other, \(E\le(k-1)|Z|\). If \(|Y|\le k-1\), then \(E\le k(k-1)\). If \(|Y|\ge k\), then \(|Z|\le k\), so \(E\le(k-1)k\). The upper bound is proved.

If the number of negative entries were even, there would be a positive number \(a\), and the product of all the other numbers would be positive. Then \(a+\prod_{\ne a}>0\), a contradiction. Hence the number of negative entries is odd.

Partition the numbers into two groups. Exactly one of the two products is negative. Let \(X=x_1x_2\cdots x_r<0\), \(Y=y_1y_2\cdots y_s>0\), and choose \(x_1<0\). Then \(x_2\cdots x_r>0\), and since the numbers are integers, \(x_2\cdots x_r\ge1\). Therefore \[X+Y=x_1(x_2\cdots x_r)+Y\le x_1+Y\le x_1+Y(x_2\cdots x_r).\] The last expression is negative by the condition applied to \(x_1\). Hence \(X+Y<0\).

Put \(t=\sqrt[2^{n+1}]{a}\). Then \(t\ne1\), \(\sqrt[2^n]{a}=t^2\), and \[x_n=2^n(t^2-1),\qquad x_{n+1}=2^{n+1}(t-1).\] Hence \[x_n-x_{n+1}=2^n(t^2-1)-2^{n+1}(t-1)=2^n(t-1)^2>0.\] Thus \(x_n>x_{n+1}\) for every \(n\), so the sequence is strictly decreasing.

For the pair \((i,j)\), the condition gives \(u_i-u_j\ge i^2-j^2\). For the pair \((j,i)\), it gives \(u_j-u_i\ge j^2-i^2\), i.e. \(u_i-u_j\le i^2-j^2\). The lower and upper bounds coincide, so \(u_i-u_j=i^2-j^2\).

Call a card uniform if both of its sides are greater than \(1\) or both are less than \(1\). Initially there is a uniform card. If a uniform card is replaced by its reciprocal, it remains uniform: large-large becomes small-small and conversely.

Now suppose a uniform card \((a,b)\) is split by the second operation. If \(a>1,b>1\), then from \(c\cdot\frac ac=a>1\), at least one of \(c,\frac ac\) is greater than \(1\); together with \(b>1\), the corresponding new card is uniform. If \(a<1,b<1\), then at least one of \(c,\frac ac\) is less than \(1\), otherwise their product would be at least \(1\); again a uniform card appears. Thus uniform cards cannot disappear completely. The desired final collection contains none, a contradiction.