Practice
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 \(4\).