Substitution Methods

Set \(x=a/b\). Then \(b/a=1/x\), and \(x+\frac1x-2=\frac{(x-1)^2}{x}\ge0\).

Then \(xyz=1\). By AM-GM, \(x+y+z\ge3\sqrt[3]{xyz}=3\), as required.

Let \(a=x^2\), \(b=y^2\), \(x,y\ge0\). We need \(x+y\le\sqrt{2(x^2+y^2)}\). Squaring gives \(2xy\le x^2+y^2\), i.e. \((x-y)^2\ge0\).

Set \(x=(b+c-a)/2\), \(y=(c+a-b)/2\), \(z=(a+b-c)/2\). Triangle inequalities give \(x,y,z>0\). Direct checking gives \(y+z=a\), \(z+x=b\), \(x+y=c\).

After substitution, \(2(ab+bc+ca)-(a^2+b^2+c^2)=4(xy+yz+zx)>0\). Hence the strict inequality holds.

Let \(a=y+z\), \(b=z+x\), \(c=x+y\). The sum becomes \(\frac12\left(\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\right)\). Inside are the pairs \(\frac{y}{x}+\frac{x}{y}\), \(\frac{z}{x}+\frac{x}{z}\), \(\frac{z}{y}+\frac{y}{z}\), each at least \(2\). Thus the sum is at least \(3\).

We have \(x+y+z=0\). Hence \(\sum a^2=\sum(1+x)^2=3+2\sum x+\sum x^2=3+\sum x^2\ge3\).

For \(x,y\ge0\), there is \(t\in[0,\pi/2]\) such that \(x=\cos t\), \(y=\sin t\). Then \(xy=\frac12\sin2t\le\frac12\).

After substitution, we need \(xy+yz+zx\le x^2+y^2+z^2\). This is equivalent to \(\frac12((x-y)^2+(y-z)^2+(z-x)^2)\ge0\).

Set \(z=1\), \(y=b\), \(x=ab\). Then \(x/y=a\), \(y/z=b\), and \(z/x=1/(ab)=c\), since \(abc=1\).

By AM-GM, \(a+b+c\ge3\sqrt[3]{abc}=3\). Equality holds at \(a=b=c=1\).

Since \(x+y+z=0\), \(ab+bc+ca=\frac13+xy+yz+zx\). But \(xy+yz+zx=-\frac12(x^2+y^2+z^2)\le0\). Hence \(ab+bc+ca\le1/3\).

For triangle sides, \(x,y,z>0\). The product equals \(8xyz>0\).

Let \(a=y+z\), \(b=z+x\), \(c=x+y\). The sum becomes \(\sum\left(1+\frac{y+z}{2x}\right)=3+\frac12\sum\frac{y+z}{x}\). As in the example, \(\sum\frac{y+z}{x}\ge6\). Hence the sum is at least \(6\).

After substitution, we need \(x^2+y^2+z^2\ge xy+yz+zx\). This is true because \(\frac12((x-y)^2+(y-z)^2+(z-x)^2)\ge0\).

Let \(x=\cos t\), \(y=\sin t\). Then \(x+y=\sqrt{2}\sin(t+\pi/4)\le\sqrt{2}\).

After substitution, the left side is \(x^2+y^2+z^2\). The right side is \(\sum(1+x)^2-3=3+2(x+y+z)+x^2+y^2+z^2-3=x^2+y^2+z^2\). The identity is proved.

Let \(u=\frac{a}{b+c-a}\), \(v=\frac{b}{c+a-b}\), \(w=\frac{c}{a+b-c}\). By Problem 6, \(u+v+w\ge3\). Then \(u^2+v^2+w^2\ge\frac{(u+v+w)^2}{3}\ge3\).

Then \(b+c-a=2x\), \(c+a-b=2y\), \(a+b-c=2z\). The first fraction is \(\frac{(y+z)^2}{4xy}\ge\frac{z}{x}\), because \((y+z)^2\ge4yz\). Similarly, the second fraction is at least \(\frac{x}{y}\), and the third at least \(\frac{y}{z}\). Hence the whole sum is at least \(\frac{z}{x}+\frac{x}{y}+\frac{y}{z}\ge3\).

For \(t\ge0\), concavity of the square root gives \(\sqrt{t}\le1+\frac{t-1}{2}\). Summing for \(a,b,c\), we get \(\sum\sqrt{a}\le3+\frac{a+b+c-3}{2}=3\). The right side in the problem is \(3+\frac12\sum(a-1)^2\ge3\), so the result follows. Equality occurs at \(a=b=c=1\).