Schur's Inequality

Schur's inequality is a classical inequality that relates three non-negative real numbers.

For non-negative real numbers \(x,\) \(y,\) and \(z,\) and for a positive real number \(t\),

\[x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y) \ge 0,\]

with equality if and only if \(x=y=z\), or if two of \(x\), \(y\), \(z\) are equal and the third is 0.

When \(t=1,\) the following special case arises:

\[x^3+y^3+z^3+3xyz \ge xy(x+y)+xz(x+z)+yz(y+z).\]

Note that the expression in the left-hand side of Schur's inequality above is symmetric in the variables \(x,\) \(y,\) and \(z.\) Then, without loss of generality, suppose \(x\ge y \ge z.\) Factoring gives

\[ (x-y)\big[x^t(x-z)-y^t(y-z)\big]+z^t(z-x)(z-y). \]

Note that \(x^t \ge y^t\) and \(x-z \ge y-z.\) Thus, the quantity inside the square brackets is non-negative. \((x-y)\) is also non-negative, so the left product is non-negative. Meanwhile, \(z^t\) is non-negative, and \((z-x)\) and \((z-y)\) are both non-positive. Thus, the right product is non-negative. Hence, the entire expression is greater than or equal to zero.\(\ _\square\)

\(a\), \(b\), and \(c\) are non-negative reals such that \(a+b+c=1\). Prove that

\[a^3 + b^3 + c^3 + 6abc \geq \dfrac {1}{4}.\]

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Multiplying out and homogenizing, we want to prove

\[\begin{align} 4a^3 + 4b^3 + 4c^3 + 24abc &\geq (a+b+c)^3\\ &= a^3 + b^3 + c^3 + 3\big(a^2 (b+c) + b^2 (c+a) + c^2 (a+b)\big) + 6abc\\ a^3 + b^3 + c^3 + 6abc &\geq a^2 (b+c) + b^2 (c+a) + c^2 (a+b). \end{align}\]

Recalling that when \(t=1\), Schur's inequality gives us \(a^3 + b^3 + c^3 + 3abc \geq a^2 (b+c) + b^2 (c+a) + c^2 (a+b)\), the inequality follows. In particular, due to the excess \(3abc\) on the LHS, equality can only be achieved in the second condition, which implies all equality solutions are permutations of \( \left(\frac {1}{2}, \frac {1}{2}, 0\right)\). \(_\square\)

Let \(a\), \(b\) and \(c\) be positive reals such that \(a+b \geq c\), \(b+c \geq a\) and \(c+a \geq b\). Prove that

\[2a^2 (b+c) + 2b^2 (c+a) + 2c^2 (a+b) \geq a^3 + b^3 + c^3 + 9abc.\]

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We do the substitution \(a=x+y\), \(b=y+z\), \(c=z+x\), noting that \(x, y, z \geq 0.\) (This is the incircle substitution including the degenerate case.) Now, the expression becomes

\[\begin{align} 4x^3 + 4y^3 + 4z^3 + 10x^2 (y+z ) + 10y^2 (z+x) + 10z^2 (x+y) + 18xyz &\geq 2x^3 + 2y^3 + 2z^3 + 12x^2 (y+z) + 12y^2 (z+x) + 12z^2 (x+y)\\ x^3 + y^3 + z^3 + 3xyz &\geq x^2 (y+z) + y^2 (z+x) + z^2 (x+y), \end{align}\]

which is true by Schur's. Equality occurs when \(x=y=z\), or if two of \(x\), \(y\) and \(z\) are equal and the other is 0. Thus, all equality cases are of the form \((t, t, t)\) or are permutations of \((2t, t, t)\). \(_\square\)