#2 AM-GM

By AM-GM, \(x+\frac{36}{x}\ge2\sqrt{x\cdot\frac{36}{x}}=12\). Equality occurs at \(x=6\).

\(\frac{a+b+c}{3}\ge\sqrt[3]{abc}=3\), hence \(a+b+c\ge9\). Equality occurs when \(a=b=c=3\).

By AM-GM, \(\sqrt[3]{abc}\le\frac{a+b+c}{3}=5\). Cubing gives \(abc\le125\).

\(\frac{x+x+y}{3}\ge\sqrt[3]{x^2y}=4\). Therefore \(2x+y\ge12\). Equality occurs at \(x=y=4\).

Using the standard AM-GM/Cauchy consequence \((a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge(1+1+1)^2=9\), and \(a+b+c=1\), we get the result.

The sum of \(1+a,1+b,1+c\) is \(9\). By AM-GM, their product is at most \(\left(\frac93\right)^3=27\). Equality occurs at \(a=b=c=2\).

By AM-GM for \(x,x,y,z\): \(\frac{x+x+y+z}{4}\ge\sqrt[4]{x^2yz}=2\). Hence \(2x+y+z\ge8\).

By AM-GM, \[\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{3}\ge\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=1.\] The result follows.

By Cauchy, \[\left(\sum\sqrt{\frac{a}{b+c}}\right)^2\ge\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}.\] Since the denominator is \(2(ab+bc+ca)\le\frac{2}{3}(a+b+c)^2=\frac23\), the sum is at least \(\sqrt{\frac32}\). This standard bound illustrates the same preparation principle.

By AM-GM, \(1+a\ge2\sqrt a\), \(1+b\ge2\sqrt b\), \(1+c\ge2\sqrt c\). Multiplying gives \((1+a)(1+b)(1+c)\ge8\sqrt{abc}=8\). Equality occurs at \(a=b=c=1\).

By AM-GM, \(\frac{x+x+x+y+z}{5}\ge\sqrt[5]{x^3yz}=\sqrt[5]{32}=2\). Hence \(3x+y+z\ge10\).

Let \(P=abcd\), \(S=a+b+c+d\). If \(P\ge16\), then \[a^2+b^2+c^2+d^2\ge4\left(\frac{S}{4}\right)^2\ge4\left(\frac{P}{8}\right)^2=\frac{P^2}{16}\ge P.\] If \(P<16\), then by AM-GM \[a^2+b^2+c^2+d^2\ge4\sqrt[4]{a^2b^2c^2d^2}=4\sqrt P.\] Since \(P<16\), \(4\sqrt P>P\). The result follows in both cases.

We have \(cd\le |cd|\le\frac{c^2+d^2}{2}=\frac{9-a^2-b^2}{2}\). It is enough to prove \(2(3+a)(3+b)\ge9-a^2-b^2\). After moving all terms to the left, we get \(a^2+2ab+b^2+6a+6b+9=(a+b+3)^2\ge0\). Hence the inequality is true.

Since \(ab+bc+ca=2\), \[a+\frac2a=a+\frac{ab+ac+bc}{a}=a+b+c+\frac{bc}{a}.\] By AM-GM, \(a+\frac{bc}{a}\ge2\sqrt{bc}\). Hence \(a+\frac2a\ge b+c+2\sqrt{bc}=(\sqrt b+\sqrt c)^2\), so \(\sqrt{a+\frac2a}\ge\sqrt b+\sqrt c\). Adding the three cyclic estimates gives the result.

Since \(|u_i|>1\), the numbers \(x_i=\frac{u_i+1}{u_i-1}\) are positive. Their product is \(1\). Also, \(\frac1{u_i-1}=\frac{x_i-1}{2}\). Thus it is enough to prove \(x_1+x_2+x_3+x_4>4\). By AM-GM, \(x_1+x_2+x_3+x_4\ge4\sqrt[4]{x_1x_2x_3x_4}=4\). Equality would require all \(x_i=1\), impossible for finite \(u_i\). Hence the sum is strictly greater than \(4\), and the required sum is positive.

After multiplying, it is enough to prove \[a^2b^2c^2+a^2b^2d^2+a^2c^2d^2+b^2c^2d^2\le1.\] Assume \(a\ge b\ge c\ge d\). By AM-GM for \(a,b,c+d\), \(ab(c+d)\le\left(\frac{a+b+c+d}{3}\right)^3=1\), so \(a^2b^2(c+d)^2\le1\). It remains to check that the left side is at most \(a^2b^2(c+d)^2\). Expanding, this reduces to \(c^2d^2(a^2+b^2)\le2a^2b^2cd\), which follows from \(cd\le b^2\) and \(cd\le a^2\). The original inequality follows.

We may assume \(x\ge y\ge z\). Let \(u=x-y\ge0\), \(v=y-z\ge0\), so \(x-z=u+v\), and the left side is \(uv(u+v)\). From \(x^2+y^2+z^2=1\), we get \((x-y)^2+(y-z)^2+(x-z)^2\le3\), hence \(u^2+v^2+(u+v)^2\le3\). For fixed \(s=u+v\), \(uv\le\frac{s^2}{4}\). Also \(u^2+v^2\ge\frac{s^2}{2}\), so \(\frac32s^2\le3\), i.e. \(s\le\sqrt2\). Therefore \(uv(u+v)\le\frac{s^3}{4}\le\frac{(\sqrt2)^3}{4}=\frac1{\sqrt2}\).

For each \(i\), \[(1+x_i^2)(1+x_{i+1}^2)-(1+x_ix_{i+1})^2=(x_i-x_{i+1})^2\ge0.\] Multiplying over all \(i\), we obtain \[\prod_{i=1}^{n}\frac{1+x_i^2}{1+x_ix_{i+1}}\ge1.\] Applying AM-GM to the \(n\) positive fractions, their sum is at least \(n\) times the \(n\)-th root of the product, hence at least \(n\).

We have \(b+c\ge2\sqrt{bc}\), hence \[4\frac{\sqrt{ab-1}}{b+c}\le2\sqrt{\frac{ab-1}{bc}}=2\sqrt{\left(a-\frac1b\right)\frac1c}\le a-\frac1b+\frac1c.\] Similarly, \[4\frac{\sqrt{bc-1}}{c+a}\le b-\frac1c+\frac1a,\qquad 4\frac{\sqrt{ca-1}}{a+b}\le c-\frac1a+\frac1b.\] Adding gives that four times the right side is at most \(a+b+c\), which is the desired inequality.

Answer: \(\frac1{76}\). Peter takes \(\frac12,\frac1{38},\ldots,\frac1{38}\). The number \(\frac12\) must be paired with \(\frac1{38}\), so the largest product is at least \(\frac1{76}\), and all other products are no larger.

Now let the numbers be arbitrary, \(x_1\ge\cdots\ge x_{20}\). Basil pairs \(x_k\) with \(x_{21-k}\). For \(k\le10\), put \(P=x_kx_{21-k}\). From \(kx_k\le x_1+\cdots+x_k=S\) and \((21-2k)x_{21-k}\le1-S\), \[P\le\frac{S(1-S)}{k(21-2k)}\le\frac1{4k(21-2k)}\le\frac1{76}.\] Thus Basil can guarantee that the maximum is at most \(\frac1{76}\). The value is proved.

Lower bound: take \(0,\frac12,\frac1{4(n-1)},\ldots,\frac1{4(n-1)}\). The number \(\frac12\) has two neighbours; at least one of them is \(\frac1{4(n-1)}\), so the maximum product is at least \(\frac1{8(n-1)}\).

For the upper bound, order \(x_1\ge\cdots\ge x_{2n}\). Arrange the numbers so that all potentially large neighbouring products have the form \(x_kx_{2n-k}\), \(1\le k\le n-1\). As in the pairing problem, if \(S=x_1+\cdots+x_k\), then \[x_kx_{2n-k}\le\frac{S(1-S)}{k(2n-2k)}\le\frac1{4k(2n-2k)}\le\frac1{8(n-1)}.\] Hence the bound is attainable for all choices and is sharp.

Since the sequence is nonincreasing, \[a_1^2+\cdots+a_{16}^2\ge4a_7^2+4a_{10}^2+4a_{13}^2+4a_{16}^2.\] Hence \[1\ge4(a_7^2+a_{10}^2+a_{13}^2+a_{16}^2)\ge(a_7+a_{10}+a_{13}+a_{16})^2,\] by RMS-AM in the last step. Thus the first sum is at most \(1\). Similarly, \(1\ge3(a_7^2+a_8^2+a_9^2)\ge(a_7+a_8+a_9)^2\), which gives the second inequality.

Subtracting gives \((a-b)(b-c)(c-a)=2\). Let \(c\) be the largest number. Then the signs imply \(a0\), \(y=c-b>0\), \(z=c-a=x+y\). Then \(xyz=2\). By AM-GM, \(4xy\le(x+y)^2=z^2\), hence \(2=xyz\le\frac{z^3}{4}\), so \(z\ge2\). If \(z=2\), equality in \(4xy\le z^2\) gives \(x=y=1\). Then \(b=a+1>1\), \(c=a+2>2\), and \(a^2b+b^2c+c^2a>b^2c>2\), contradicting the first equality. Therefore \(z>2\), i.e. \(c-a>2\).

Add the first two inequalities: \(2x+2y>x^2+y^2+2z^2\). Rearranging, \[0>(x-1)^2+(y-1)^2+2(z^2-1).\] Since the first two terms are nonnegative, \(z^2<1\), hence \(z<1\). Similarly, adding the other pairs gives \(x<1\) and \(y<1\).

Multiplying the three factors by \(\frac ba,\frac cb,\frac ac\), we get \[\left(b-\frac1a\right)\left(c-\frac1b\right)\left(a-\frac1c\right)=1.\] All three factors are positive. By AM-GM, \[\left(b-\frac1a\right)^2+\left(c-\frac1b\right)^2+\left(a-\frac1c\right)^2\ge3.\] Expanding gives \[a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}\ge3+2\left(\frac ba+\frac cb+\frac ac\right).\] Also, \(\frac ba+\frac cb+\frac ac\ge3\). Hence the right side is at least \(6+\frac ba+\frac cb+\frac ac\). Expanding the left side of the desired inequality gives \(a^2+b^2+c^2+\frac1{a^2}+\frac1{b^2}+\frac1{c^2}-6\). The result follows.