Inequalities with Constraints

By Cauchy, \(36=(a+b+c)^2\le3(a^2+b^2+c^2)\). Therefore \(\sum a^2\ge12\). Equality occurs at \(a=b=c=2\), so the minimum is \(12\).

By AM-GM, \(\sqrt[3]{abc}\le(a+b+c)/3=3\). Hence \(abc\le27\). Equality holds at \(a=b=c=3\).

By AM-GM, \(a+b+c\ge3\sqrt[3]{abc}=6\). Equality occurs at \(a=b=c=2\).

Since \((a+b+c)^2\ge3(ab+bc+ca)\), with \(a+b+c=1\) we get \(1\ge3(ab+bc+ca)\). Equality holds at \(a=b=c=1/3\).

By Cauchy, \(9=(1+1+1)^2\le(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)=5\sum1/a\). Hence \(\sum1/a\ge9/5\).

Let \(p=a+b+c\), \(q=ab+bc+ca=12\). Then \(p^2\ge3q=36\), so \(p\ge6\). Equality occurs at \(a=b=c=2\).

By Jensen, \(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}3\le\sqrt{\frac{x+y+z}{3}}=2\). Thus the sum is at most \(6\). Equality holds at \(x=y=z=4\).

The function \(\ln(1+x)\) is concave for \(x\ge0\). Hence \(\frac13\sum\ln(1+a)\le\ln(1+(a+b+c)/3)=\ln2\). Therefore \(\ln\prod(1+a)\le\ln8\), and the product is at most \(8\).

The function \(x^3\) is convex on \(x\ge0\). By Jensen, \(\frac{\sum a^3}{3}\ge\left(\frac{S}{3}\right)^3\). Therefore \(\sum a^3\ge S^3/9\). Equality holds at \(a=b=c=S/3\).

Since \(0\le a,b,c\le3\), we have \(a^2\le3a\), \(b^2\le3b\), \(c^2\le3c\). Therefore \(\sum a^2\le3(a+b+c)=9\). Equality occurs at \((3,0,0)\) and permutations.

By AM-GM, \(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}=3\). Equality holds when \(ab=bc=ca=1\), i.e. \(a=b=c=1\).

Expanding, \((1+a)(1+b)(1+c)=1+(a+b+c)+(ab+bc+ca)+abc\). Since \(abc=1\), \(a+b+c\ge3\), and \(ab+bc+ca\ge3\), the product is at least \(1+3+3+1=8\).

By Jensen for the concave function \(f(x)=x/(1+x)\), \(\frac13\sum f(a)\le f(1)=1/2\). Hence the sum is at most \(3/2\). Equality holds at \(a=b=c=1\).

By Jensen for the convex function \(f(x)=1/(1+x)\), \(\frac13\sum f(a)\ge f(1)=1/2\). Therefore the sum is at least \(3/2\).

\(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=9-4=5\). The answer is \(5\).

We have \((a+b)(b+c)(c+a)=(3-c)(3-a)(3-b)\). The sum of the factors is \((3-a)+(3-b)+(3-c)=6\). By AM-GM, their product is at most \((6/3)^3=8\). Equality holds at \(a=b=c=1\).

Let \(p=a+b+c\), \(q=3\), \(r=abc\). From \(p^2\ge3q=9\), we get \(p\ge3\). Also \(q^2\ge3pr\), i.e. \(9\ge3pr\), so \(r\le3/p\le1\). Therefore \(abc\le1\). Equality holds at \(a=b=c=1\).

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

By Cauchy, \[\sum\frac{a^2}{1+a}\ge\frac{(a+b+c)^2}{(1+a)+(1+b)+(1+c)}=\frac{9}{6}=\frac32.\] Equality holds at \(a=b=c=1\).

Since \(a+b+c=3\), \(\sum a^2+ab+bc+ca=(a+b+c)^2-(ab+bc+ca)=9-q\). We need \(9-q\ge6r\). From \(q\le3\), the left side is at least \(6\), and by AM-GM \(r=abc\le1\), so \(6r\le6\). Therefore \(9-q\ge6\ge6r\). Equality holds at \(a=b=c=1\).