Chapter
Theory
The UVW method works with symmetric inequalities in three variables. Instead of \(a,b,c\), we look at \(p=a+b+c\), \(q=ab+bc+ca\), \(r=abc\). Often, after fixing \(p,q\), the expression depends on \(r\) in a simple way, and the hardest case occurs on the boundary or when two variables are equal.
Main notation: \(p=a+b+c\), \(q=ab+bc+ca\), \(r=abc\). Useful formulas:
\[a^2+b^2+c^2=p^2-2q,\qquad a^3+b^3+c^3=p^3-3pq+3r.\]
Also \(\sum_{\mathrm{sym}}a^2b=pq-3r\), where \(\sum_{\mathrm{sym}}a^2b=a^2b+a^2c+b^2a+b^2c+c^2a+c^2b\). Schur degree \(3\): \[p^3-4pq+9r\ge0.\]
Use the method when the inequality is symmetric in \(a,b,c\), equality is expected at \(a=b=c\) or at two equal variables, and the expression can be rewritten through \(p,q,r\). It is especially useful for degrees \(3\) and \(4\), Schur-type inequalities, and fixed-sum problems.
Check whether the problem is unchanged by permuting \(a,b,c\). Then identify the degree and try to express the sums through \(p,q,r\). If after fixing \(p,q\) the dependence on \(r\) is linear, it is natural to check the cases \(b=c\) and \(c=0\).
Do not apply UVW to a nonsymmetric expression. Do not confuse the symmetric sum \(\sum_{\mathrm{sym}}\) with a cyclic sum. Another mistake is checking only \(a=b=c\): a proof usually requires the two-equal case or the boundary.
1. Is the inequality symmetric? 2. Which \(p,q,r\) appear? 3. Are \(p\) or \(q\) fixed? 4. Can Schur be applied? 5. What do the cases \(b=c\) and \(c=0\) give? 6. Does the equality case match the answer?
Examples
Problem. Express \(a^2+b^2+c^2\) and \(a^3+b^3+c^3\) through \(p,q,r\).
From \(p^2=a^2+b^2+c^2+2q\), we get \(\sum a^2=p^2-2q\). Also \(p^3=\sum a^3+3\sum_{\mathrm{sym}}a^2b+6r\), and \(\sum_{\mathrm{sym}}a^2b=pq-3r\), so \(\sum a^3=p^3-3pq+3r\).
Problem. Prove \(\sum_{\mathrm{sym}}a^2b=pq-3r\).
Expand \(pq=(a+b+c)(ab+bc+ca)\). We get all six terms \(a^2b,a^2c,\ldots,c^2b\), plus \(3abc\). Hence \(\sum_{\mathrm{sym}}a^2b=pq-3r\).
Problem. Prove \(\sum a^3+3abc\ge\sum_{\mathrm{sym}}a^2b\).
The left side minus the right side equals \((p^3-3pq+3r)+3r-(pq-3r)=p^3-4pq+9r\). This is Schur degree \(3\), hence it is nonnegative.
Problem. Check Schur degree \(3\) in the case \(b=c=1\), \(a=t\).
The inequality becomes \(t^3+2+3t\ge 2t^2+2t+2\), or \(t(t-1)^2\ge0\). This is a typical two-equal-variable check.
Problem. Prove \(p^2\ge3q\).
\(p^2-3q=a^2+b^2+c^2-ab-bc-ca=\frac12((a-b)^2+(b-c)^2+(c-a)^2)\ge0\).
Problem. If \(a+b+c=p\), prove \(abc\le\frac{p^3}{27}\).
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Therefore \(abc\le(p/3)^3\). In UVW language, this agrees with the fact that for fixed \(p\), the maximum of the symmetric product occurs at \(a=b=c\).
Problem. Prove \(\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b\).
This is Schur degree \(4\). By UVW, it is enough to check the two-equal case and the boundary. For \(b=c=1\), \(a=t\), the difference is \(t^2(t-1)^2\ge0\); on the boundary we get an obvious sum of squares. Hence the inequality holds.
Problem. Why can UVW not be applied directly to \(a^3b+b^3c+c^3a\le a^4+b^4+c^4\)?
The right side is symmetric, but the left side is cyclic and changes under permutations of \(a,b,c\). Therefore UVW does not apply directly; this problem is better handled by rearrangement or by summing over all permutations.
Problems
Let \(p=a+b+c\), \(q=ab+bc+ca\). Prove \[a^2+b^2+c^2=p^2-2q.\]
Hint. Expand \((a+b+c)^2\).
\(p^2=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=\sum a^2+2q\). Hence \(\sum a^2=p^2-2q\).
Let \(p=a+b+c\), \(q=ab+bc+ca\), \(r=abc\). Prove \[a^3+b^3+c^3=p^3-3pq+3r.\]
Hint. Use \(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\).
By the identity, \(\sum a^3-3r=p((p^2-2q)-q)=p(p^2-3q)=p^3-3pq\). Therefore \(\sum a^3=p^3-3pq+3r\).
Prove \[\sum_{\mathrm{sym}}a^2b=pq-3r.\]
Hint. Expand \((a+b+c)(ab+bc+ca)\).
Expanding \(pq\), we get the six terms \(a^2b,a^2c,b^2a,b^2c,c^2a,c^2b\) and three copies of \(abc\). Thus \(pq=\sum_{\mathrm{sym}}a^2b+3r\), which gives the formula.
Prove \[p^2-3q=\frac12\left((a-b)^2+(b-c)^2+(c-a)^2\right).\]
Hint. Expand the right side and compare it with \(p^2-3q\).
The right side equals \(\frac12(2a^2+2b^2+2c^2-2ab-2bc-2ca)=a^2+b^2+c^2-q\). Also \(p^2-3q=(a^2+b^2+c^2+2q)-3q=a^2+b^2+c^2-q\).
Prove that the inequality \(\sum a^3+3abc\ge\sum_{\mathrm{sym}}a^2b\) is equivalent to \[p^3-4pq+9r\ge0.\]
Hint. Substitute the formulas for \(\sum a^3\) and \(\sum_{\mathrm{sym}}a^2b\).
The difference between the left and right sides is \((p^3-3pq+3r)+3r-(pq-3r)=p^3-4pq+9r\). Therefore the original inequality is equivalent to the nonnegativity of this expression.
Prove for \(a,b,c\ge0\): \[a^2+b^2+c^2\ge ab+bc+ca.\]
Hint. Use the formula for \(p^2-3q\).
The inequality is equivalent to \(p^2-2q\ge q\), i.e. \(p^2-3q\ge0\). But \(p^2-3q=\frac12((a-b)^2+(b-c)^2+(c-a)^2)\ge0\).
Let \(a,b,c\ge0\) and \(a+b+c=p\). Prove \[ab+bc+ca\le\frac{p^2}{3}.\]
Hint. Use \(p^2\ge3q\).
By \(p^2-3q=\frac12\sum(a-b)^2\ge0\), we get \(q\le\frac{p^2}{3}\). Equality is possible only at \(a=b=c\).
Let \(a,b,c\ge0\) and \(a+b+c=p\). Prove \[abc\le\frac{p^3}{27}.\]
Hint. Use AM-GM, or check that the maximum occurs at \(a=b=c\).
By AM-GM, \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\). Hence \(\sqrt[3]{abc}\le p/3\), so \(abc\le p^3/27\). Equality holds at \(a=b=c=p/3\).
Let \(b=c=1\), \(a=t\ge0\). Check the inequality \[\sum a^3+3abc\ge\sum_{\mathrm{sym}}a^2b.\]
Hint. Substitute \(a=t\), \(b=c=1\), and factor the difference.
The left side is \(t^3+2+3t\). The right side is \(2t^2+2t+2\). The difference is \(t^3-2t^2+t=t(t-1)^2\ge0\).
Prove for \(a,b,c\ge0\): \[\sum a^3+3abc\ge\sum_{\mathrm{sym}}a^2b.\]
Hint. Use the form \(p^3-4pq+9r\ge0\), i.e. Schur degree \(3\).
By Problem 5, the difference between the left and right sides is \(p^3-4pq+9r\). By Schur degree \(3\), this expression is nonnegative for \(a,b,c\ge0\). Hence the inequality follows.
Prove the formula \[a^4+b^4+c^4=p^4-4p^2q+2q^2+4pr.\]
Hint. Use \(a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)\).
We have \(\sum a^2=p^2-2q\). Also \(a^2b^2+b^2c^2+c^2a^2=q^2-2pr\). Thus \[\sum a^4=(p^2-2q)^2-2(q^2-2pr)=p^4-4p^2q+2q^2+4pr.\]
Prove that \[a^4+b^4+c^4-abc(a+b+c)=p^4-4p^2q+2q^2+3pr.\]
Hint. Use the previous formula and \(abc(a+b+c)=pr\).
By the previous problem, \(\sum a^4=p^4-4p^2q+2q^2+4pr\). Subtracting \(abc(a+b+c)=pr\), we obtain \(p^4-4p^2q+2q^2+3pr\).
Check the inequality \[a^4+b^4+c^4\ge abc(a+b+c)\] in the case \(b=c=1\), \(a=t\ge0\).
Hint. The difference should have the factor \((t-1)^2\).
After substitution, we get \(t^4+2\ge t(t+2)\). The difference is \(t^4-t^2-2t+2=(t-1)^2(t^2+2t+2)\ge0\).
Prove for \(a,b,c\ge0\): \[a^4+b^4+c^4\ge abc(a+b+c).\]
Hint. This symmetric inequality is linear in \(r\) for fixed \(p,q\). Check \(b=c\) and the boundary.
By UVW, it is enough to check the two-equal case and the boundary. If \(b=c=1\), \(a=t\), the difference is \((t-1)^2(t^2+2t+2)\ge0\). If one variable is \(0\), the right side is \(0\), while the left side is nonnegative. Hence the inequality holds.
Let \(b=c=1\), \(a=t\ge0\). Check \[\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b.\]
Hint. Write both sides explicitly and simplify.
The left side is \(t^4+2+t(t+2)=t^4+t^2+2t+2\). The right side is \(2t^3+2t+2\). The difference is \(t^4-2t^3+t^2=t^2(t-1)^2\ge0\).
Prove for \(a,b,c\ge0\): \[\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b.\]
Hint. This is Schur degree \(4\); use the UVW reduction to \(b=c\) and the boundary.
The inequality is symmetric and of Schur type in degree \(4\). By UVW, it is enough to check \(b=c\) and the boundary. For \(b=c=1\), \(a=t\), the difference is \(t^2(t-1)^2\ge0\). On the boundary, say \(c=0\), we get \(a^4+b^4\ge a^3b+ab^3\), i.e. \((a-b)^2(a^2+ab+b^2)\ge0\). Hence the inequality is proved.
Prove for \(a,b,c\ge0\): \[(ab+bc+ca)^2\ge3abc(a+b+c).\]
Hint. Apply \(x^2+y^2+z^2\ge xy+yz+zx\) to \(x=ab\), \(y=bc\), \(z=ca\).
Let \(x=ab\), \(y=bc\), \(z=ca\). Then \((x+y+z)^2\ge3(xy+yz+zx)\). The left side is \(q^2\), and \(xy+yz+zx=abc(a+b+c)=pr\). Therefore \(q^2\ge3pr\).
Let \(a,b,c\ge0\) and \(a+b+c=1\). Prove \[a^3+b^3+c^3+6abc\ge ab+bc+ca.\]
Hint. Rewrite \(\sum a^3\) as \(1-3q+3r\).
Since \(p=1\), \(\sum a^3=1-3q+3r\). The left side minus the right side equals \(1-4q+9r\). This is Schur degree \(3\) with \(p=1\): \(p^3-4pq+9r\ge0\). Hence the inequality holds.
Let \(a,b,c\ge0\) and \(a+b+c=p\). Using the two-equal-variables idea, prove \(abc\le\frac{p^3}{27}\).
Hint. Check \(b=c=x\), \(a=p-2x\), and factor \(\frac{p^3}{27}-(p-2x)x^2\).
By UVW, for fixed \(p\), the extremum of \(r=abc\) occurs when two variables are equal or on the boundary. On the boundary \(r=0\). Let \(b=c=x\), \(a=p-2x\). Then \[ \frac{p^3}{27}-(p-2x)x^2=\frac{(p-3x)^2(p+6x)}{27}\ge0. \] Hence \(abc\le p^3/27\).
Let \(a,b,c\ge0\) and \(a+b+c=3\). Prove \[a^4+b^4+c^4+3abc\ge\sum_{\mathrm{sym}}a^3b.\]
Hint. This is Schur degree \(4\) with \(p=3\).
The general Schur degree \(4\) is \(\sum a^4+abc(a+b+c)\ge\sum_{\mathrm{sym}}a^3b\). Under \(a+b+c=3\), we have \(abc(a+b+c)=3abc\). Thus the required inequality follows directly from Schur degree \(4\).
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