Cauchy-Schwarz

By Cauchy for \((a,b)\) and \((c,d)\), \((a^2+b^2)(c^2+d^2)\ge(ac+bd)^2\). After expansion, the difference is \((ad-bc)^2\ge0\).

By Engel form, \(\frac{x^2}{p}+\frac{y^2}{q}\ge\frac{(x+y)^2}{p+q}\). Equality occurs when \(\frac{x}{p}=\frac{y}{q}\).

By Cauchy, \[9=(1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)=12\left(\frac1a+\frac1b+\frac1c\right).\] The result follows.

\[\sum\frac{a}{b+c}=\sum\frac{a^2}{a(b+c)}\ge\frac{(a+b+c)^2}{a(b+c)+b(c+a)+c(a+b)}=\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac32,\] since \((a+b+c)^2\ge3(ab+bc+ca)\).

By Cauchy, \[\sum\frac{x^2}{x+y}\ge\frac{(x+y+z)^2}{(x+y)+(y+z)+(z+x)}=\frac{x+y+z}{2}.\]

By Cauchy, the left side is at least \[\frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}.\] Since \(2(x^2+y^2+z^2)+xy+yz+zx\le2(x+y+z)^2\), the left side is at least \(\frac12\).

By Cauchy, \[\sum_{i=1}^{n}\frac{1^2}{1+x_i}\ge\frac{(1+\cdots+1)^2}{(1+x_1)+\cdots+(1+x_n)}=\frac{n^2}{n+\sum x_i}.\]

By Cauchy, \[\frac{4}{a}+\frac{9}{b}+\frac{16}{c}\ge\frac{(2+3+4)^2}{a+b+c}=\frac{81}{6}=\frac{27}{2}.\] Equality occurs when \(\frac{2}{a}=\frac{3}{b}=\frac{4}{c}\), i.e. \(a:b:c=2:3:4\). This is attainable with sum \(6\), so the minimum is \(\frac{27}{2}\).

By Cauchy, \[\sum\frac{a^2}{a+2b}\ge\frac{(a+b+c)^2}{(a+2b)+(b+2c)+(c+2a)}=\frac{(a+b+c)^2}{3(a+b+c)}=\frac{a+b+c}{3}.\]

By Cauchy, the left side is at least \[\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}.\] Since \(2(a^2+b^2+c^2)+ab+bc+ca\le2(a+b+c)^2\), we get the lower bound \(\frac12\).

By Cauchy, \[\sum\frac{a^2}{b+2c}\ge\frac{(a+b+c)^2}{(b+2c)+(c+2a)+(a+2b)}=\frac{(a+b+c)^2}{3(a+b+c)}=\frac{a+b+c}{3}.\]

By Cauchy, \[\frac1x+\frac4y+\frac9z\ge\frac{(1+2+3)^2}{x+y+z}=\frac{36}{10}=\frac{18}{5}.\] Equality occurs when \(1/x=2/y=3/z\), i.e. \(x:y:z=1:2:3\). This is attainable with sum \(10\), so the minimum is \(\frac{18}{5}\).

By Cauchy, the left side is at least \[\frac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}.\] This denominator is at most \(2(a+b+c)^2\), so the left side is at least \(\frac12\). The condition \(abc=1\) shows that the problem may look multiplicative, but the method is homogeneous.

By Cauchy, the left side is at least \[\frac{(a+b+c)^2}{(a+b+c)+(a+2b+c)+(a+b+3c)}=\frac{(a+b+c)^2}{3a+4b+5c}.\]

By Cauchy, \[(\sqrt{u+v}+\sqrt{v+w}+\sqrt{w+u})^2\le3((u+v)+(v+w)+(w+u))=6(u+v+w)=36.\] Hence the sum of roots is at most \(6\).

For each term, \[\frac{a^3}{a^2+b+c}=a-\frac{a(b+c)}{a^2+b+c}\ge a-\frac{a(b+c)}{2a\sqrt{b+c}}=a-\frac{\sqrt{b+c}}2.\] Adding the four estimates and using \(a+b+c+d=10\) gives the result.

As in the previous problem, \(\frac{a^3}{a^2+b+c}\ge a-\frac{\sqrt{b+c}}2\), and similarly for the other terms. Thus the left side is at least \[8-\frac12(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a}).\] It remains to show that the root sum is at most \(8\). By Cauchy, \[(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+d}+\sqrt{d+a})^2\le4\cdot2(a+b+c+d)=64.\] Hence the root sum is at most \(8\), and the left side is at least \(4\).

\[\frac{a}{b^4+2b}=\frac{a}{b}-\frac{2ab^2}{b^3+2}\ge\frac{a}{b}-\frac{2ab^2}{3b}=\frac{a}{b}-\frac23ab,\] because \(b^3+2=b^3+1+1\ge3b\). Adding cyclically, \[\sum\frac{a}{b^4+2b}\ge\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\frac23(ab+bc+ca).\] The first bracket is at least \(3\) by AM-GM. Also, \(a+b+c=3\) implies \(ab+bc+ca\le3\). Thus the right side is at least \(3-\frac23\cdot3=1\).

By Cauchy, the left side is at least \[\frac{(x^2+y^2+z^2)^2}{2(x^2+y^2+z^2)+xy+yz+zx}.\] Since \(xy+yz+zx\le x^2+y^2+z^2\), the denominator is at most \(3(x^2+y^2+z^2)\). This gives the required bound \(\frac{x^2+y^2+z^2}{3}\).

By Cauchy, \[\sum\frac{a^2}{2a+b+c}\ge\frac{(a+b+c)^2}{(2a+b+c)+(2b+c+a)+(2c+a+b)}=\frac{(a+b+c)^2}{4(a+b+c)}=\frac{a+b+c}{4}.\]