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

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

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

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

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 $(2t, t, t)$. $_\square$