# Inequality mania!

Let $$x,y,z \ge 0$$ where $$x+y+z=1$$. Prove that

$\large{x^3+y^3+z^3+6xyz \ge \dfrac{1}{4} }.$

Note by Department 8
2 years, 5 months ago

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$$\frac{1}{4}=\frac{(x+y+z)^3}{4}$$

$$x^3+y^3+z^3+6xyz\geq\frac{(x+y+z)^3}{4}$$

$$4x^3+4y^3+4z^3+24xyz\geq x^3+y^3+z^3+3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2+6xyz$$

$$3x^3+3y^3+3z^3+18xyz\geq3x^2y+3xy^2+3y^2z+3yz^2+3z^2x+3zx^2$$

$$x^3+y^3+z^3+6xyz\geq x^2y+xy^2+y^2z+yz^2+z^2x+zx^2$$

Which is true by Schur's. Equality when two of $$x, y, z$$ equal $$\frac{1}{2}$$ and the third equals $$0$$. Note that the original inequality would still hold if it were $$x^3+y^3+z^3+3.75xyz$$ instead of $$x^3+y^3+z^3+6xyz$$

- 2 years, 5 months ago

Wrote very nicely+1

- 2 years, 5 months ago