#7 Homogeneous Inequalities

The first expression is multiplied by \(t^2\), so its degree is \(2\). The second does not change, so its degree is \(0\). The third is multiplied by \(t\), so its degree is \(1\).

The first is homogeneous of degree \(2\), so normalization is allowed. The second is not homogeneous: the degrees are \(2\) and \(0\), so normalization is not allowed. The third has degree \(0\), so normalization is allowed for positive variables.

The inequality is homogeneous. With \(a+b+c=1\), we need \(a^2+b^2+c^2\ge1/3\). This follows from \((a+b+c)^2\le3(a^2+b^2+c^2)\). Returning to the original scale gives the result.

Set \(x=a/\sqrt[3]{abc}\), \(y=b/\sqrt[3]{abc}\), \(z=c/\sqrt[3]{abc}\). Then \(xyz=1\). We need \(x+y+z\ge3\), which follows from AM-GM.

Since \(a+b+c=1\), it is enough to prove \(ab+bc+ca\le\frac{(a+b+c)^2}{3}\), i.e. \(3q\le p^2\). This follows from \(p^2-3q=\frac12\sum(a-b)^2\ge0\).

Let \(x=a/b\). The inequality becomes \(x+\frac1x\ge2\), which follows from \((x-1)^2\ge0\) after multiplying by \(x>0\).

The expression has degree \(0\). By AM-GM, \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\sqrt[3]{\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}}=3.\]

By AM-GM, \[\sum\frac{a^2}{b^2}\ge3\sqrt[3]{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=3.\]

By Cauchy, \[\sum\frac{a^2}{a+b}\ge\frac{(a+b+c)^2}{(a+b)+(b+c)+(c+a)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}.\]

By Cauchy, \[\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}.\] Since \((a+b+c)^2\ge3(ab+bc+ca)\), we get the lower bound \(\frac32\).

By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Since the sum is \(1\), \(\sqrt[3]{abc}\le1/3\), hence \(abc\le1/27\). Homogeneous form: \(abc\le\frac{(a+b+c)^3}{27}\).

By AM-GM, \(\frac{a^2+b^2+c^2}{3}\ge\sqrt[3]{a^2b^2c^2}=1\). Therefore \(\sum a^2\ge3\).

By Cauchy, \((1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\). Therefore the product is at least \(9\).

By Cauchy, \((a+b+c)^2\le3(a^2+b^2+c^2)\). Since \(a+b+c=3\), we get \(9\le3\sum a^2\), so \(\sum a^2\ge3\).

By Cauchy, \[\sum\frac{a^2}{b+c}\ge\frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}.\]

If \(a+b+c=0\), the result is obvious. Otherwise set \(S=a+b+c\), \(x=a/S\), \(y=b/S\), \(z=c/S\). Then \(x+y+z=1\), so \(\frac{a^2+b^2+c^2}{S^2}\ge1/3\). Multiply by \(S^2\).

This is a symmetric homogeneous inequality of degree \(4\). By the UVW form already proved, it is enough to check \(b=c=1\), \(a=t\): \(t^4+2\ge t(t+2)\), equivalent to \((t-1)^2(t^2+2t+2)\ge0\). On the boundary, the right side is \(0\). Hence the inequality holds.

The expression has degree \(0\). Divide \(a,b,c\) by \(\sqrt[3]{abc}\); the product becomes \(1\), and the expression does not change. With \(abc=1\), we need \(a^2+b^2+c^2\ge3\), which follows from AM-GM.

Let \(q=ab+bc+ca\). Since \(p^2=\sum a^2+2q=1\), we have \(\sum a^2+q=1-q\). Also \(q\le1/3\), so \(1-q\ge2/3\). The homogeneous version is \[\sum a^2+ab+bc+ca\ge\frac{2}{3}(a+b+c)^2.\]

The expression has degree \(0\), so normalize \(abc=1\). Then \(\sqrt[3]{a^2b^2c^2}=1\). We get \[\frac{\sum a^2}{q}+q.\] Since \(\sum a^2\ge q\), the first fraction is at least \(1\). Also \(q=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\). Hence the whole sum is at least \(1+3=4\).