Mixed Inequality Sets

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 \((a+b+c)^2\ge3(ab+bc+ca)\), the lower bound is \(3/2\).

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

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

From \(abc=1\), AM-GM gives \(a+b+c\ge3\). Also \(ab\cdot bc\cdot ca=(abc)^2=1\), so \(ab+bc+ca\ge3\). Adding gives \(6\).

We have \(a+b\ge2\sqrt{ab}\), \(b+c\ge2\sqrt{bc}\), \(c+a\ge2\sqrt{ca}\). Multiplying gives \((a+b)(b+c)(c+a)\ge8abc\).

It is enough to prove for \(x,y,z>0\) that \(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\ge1\). With common denominator \((x+y)(y+z)(z+x)\), the numerator of the difference is \(xyz+x^2y+xz^2+y^2z\), which is positive. Now take \(x=a^2\), \(y=b^2\), \(z=c^2\).

By Cauchy, \[\sum\frac{a^2}{a+b}\ge\frac{(a+b+c)^2}{2(a+b+c)}=\frac12.\]

For \(x,y>0\), \[3x^3-(2x-y)(x^2+xy+y^2)=(x-y)^2(x+y)\ge0.\] Hence \(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\). Summing for \((a,b)\), \((b,c)\), \((c,a)\), gives \(\frac{2a-b+2b-c+2c-a}{3}=\frac{a+b+c}{3}\).

By Cauchy, \[\sum\frac1{a+2}\ge\frac{(1+1+1)^2}{a+b+c+6}=\frac9{9}=1.\]

By Cauchy, \[\sum\frac{a^2}{(a+b)(a+c)}\ge\frac{(a+b+c)^2}{\sum(a+b)(a+c)}.\] The denominator is \(a^2+b^2+c^2+3(ab+bc+ca)\). Thus it remains to check \[4(a+b+c)^2\ge3(a^2+b^2+c^2+3ab+3bc+3ca),\] which is equivalent to \(a^2+b^2+c^2\ge ab+bc+ca\).

For \(x\ge0\), \[\frac1{1+x^2}-\frac{2-x}{2}=\frac{x(x-1)^2}{2(1+x^2)}\ge0.\] Summing for \(x=a,b,c\), we get \[\sum\frac1{1+a^2}\ge\frac{6-(a+b+c)}2=\frac32.\]

Since \[x^2+xy+y^2-\frac34(x+y)^2=\frac14(x-y)^2\ge0,\] we have \(\sqrt{x^2+xy+y^2}\ge\frac{\sqrt{3}}2(x+y)\). Summing over \((a,b)\), \((b,c)\), \((c,a)\), gives \(\frac{\sqrt{3}}2\cdot2(a+b+c)=3\sqrt{3}\).

By Holder, \[\sum\frac{a^3}{b^2+bc+c^2}\ge\frac{(a+b+c)^3}{2(a^2+b^2+c^2)+ab+bc+ca}.\] Since \(3(a+b+c)^2\ge2(a^2+b^2+c^2)+ab+bc+ca\), the lower bound is \((a+b+c)/3\).

By Schur, \(\sum a(a-b)(a-c)\ge0\). Expanding gives \[a^3+b^3+c^3+3abc-\sum_{\mathrm{sym}}a^2b\ge0,\] which is exactly the claim.

By Cauchy, \[\sum\frac{a^2}{b^2+c^2+a(b+c)}\ge\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+2(ab+bc+ca)}=\frac{(a+b+c)^2}{2(a+b+c)^2}=\frac12.\]

The condition gives \(A>0\). Also \(C=P(0)>0\), and the discriminant \(B^2-4AC<0\). For \(C x^2+Bx+A\), the leading coefficient is positive and the discriminant is the same: \(B^2-4CA<0\). Hence it is positive for all real \(x\).

For \(u,v>0\), compare the root with \(u+v\). If \(v\ge u\), then \(\sqrt{u^2+3v^2}\ge u+v\). If \(v\le u\), then \(\sqrt{u^2+3v^2}\le u+v\), but \(v-u\le0\), so the direction reverses after multiplication. In both cases \[(v-u)\sqrt{u^2+3v^2}\ge(v-u)(u+v)=v^2-u^2.\] Applying this to \(u=x_i\), \(v=x_{i+1}\), the sum is at least \(\sum(x_{i+1}^2-x_i^2)=0\).

By Cauchy, \[(\sum a^3)^2=(\sum a\cdot a^2)^2\le(\sum a^2)(\sum a^4).\] Also, \(\sum a^4\le(\sum a^2)^2\). Therefore \((\sum a^3)^2\le(\sum a^2)^3\).

By Schur, \[\sum a^3(a-b)(a-c)\ge0.\] Expanding gives \[\sum a^5+abc(a^2+b^2+c^2)-\sum_{\mathrm{sym}}a^4b\ge0,\] which is equivalent to the required inequality.

Order the numbers as \(y_1