AMGM or Holder

anantmudgal09 wrote: Let $x,y,z$ be non-negative real numbers such that $xyz=1$. Prove that $$(x^3+2y)(y^3+2z)(z^3+2x) \ge 27.$$ $(x^3+2y)(y^3+2z)(z^3+2x) \ge 3\sqrt[3]{x^3y^2}\cdot3\sqrt[3]{y^3z^2}\cdot3\sqrt[3]{z^3x^2}=27.$

Simply write 2y as y + y . Same with 2z and 2x. Now apply AM-GM Inequality on each term within brackets on the left hand side to get the desired result.

$x^3+y+y\ge 3x(y^{\frac{2}{3}}$ Like this multuply all and get $P(x,y,z)\ge 27xyz=27.1=27$

WizardMath wrote: AMGM or Holder Holder $$(x^3+y+y)(y^3+z+z)(z^3+x+x)\ge (xyz+\sqrt[3]{xyz}+\sqrt[3]{xyz})^3=3^3$$

Observe that equality holds when $x=y=z=1$. So, directly applying AM-GM on $x^3+2y$ won't help. But, we can write $2y$ as $y+y$. Hence, by AM-GM, $LHS=(x^3+y+y)(y^3+z+z)(z^3+x+x)\geq27(\sqrt[3]{x^3y^2\times y^3z^2\times z^3x^2})=27\sqrt[3]{(xyz)^5}=27$

anantmudgal09 wrote: Let $x,y,z$ be non-negative real numbers such that $xyz=1$. Prove that $$(x^3+2y)(y^3+2z)(z^3+2x) \ge 27.$$ Because: $$(x^3+2y)(y^3+2z)(z^3+2x)=\left(\frac{x^3}{y}+2\right)\left(\frac{y^3}{z}+2\right)\left(\frac{z^3}{x}+2\right) \geq 9\left(y\sqrt{\frac{x^3}{z}}+z\sqrt{\frac{y^3}{x}}+x\sqrt{\frac{z^3}{y}}\right) \geq 27$$

Quite old but still surprising that it was a p5 $x^3+2y = x^3 + y + y \ge 3 \sqrt[3]{x^3 y^2}$ $y^3 + 2z \ge 3 \sqrt[3]{y^3z^2}$ $z^3 + 2x \ge 3 \sqrt[3]{z^3 x^2}$ Multiplying all we have $(x^3+2y)(y^3+2z)(z^3+2x) \ge 27 \sqrt[3]{x^5y^5z^5} = 27$ Equality occurring when $x=y=z=1$