ACGNmath wrote: Prove the inequality $$(x^3+2y^2+3z)(4y^3+5z^2+6x)(7z^3+8x^2+9y)\geq720(xy+yz+xz)$$for $x, y, z \geq 1$. Proposed by K. Kokhas Because $$x^3+2y^2+3z\geq2y^2+6,$$$$4y^3+5z^2+6x\geq5z^2+10$$and $$7z^3+8x^2+9y\geq8x^2+16.$$

i would do like this:$x^3+2y^2+3z\ge{x^3+y^2+z+y^2+z+1}$ $4y^3+5z^2+6x=\frac{5}{2}(y^3+z^2+x)+\frac{3}{2}(y^3+z^2+x)+z^2+2x\ge{\frac{5}{2}(y^3+z^2+x+z^2+x+1)}$ and $ 7z^3+8x^2+9y\ge{4(z^3+x^2+y+x^2+y+1)}$ Apply Holder twice to get $LHS\ge{10(\sqrt[3]{\Pi{(x^3+y^2+z)}}+\sqrt[3]{\Pi{(x^2+y+1)}})^3}\ge{10((x^2+y^2+z^2)+(x+y+z))^3}\ge{10(\sigma+3)^3}$ Where $\sigma=\Sigma xy$ We are left to prove that $(\sigma+3)^3\ge{72\sigma}$, which is true since $(\sigma+3)^3\ge{8(\sqrt{3\sigma})^3}\ge{72\sigma}$ since $\sigma\ge{3}$

$x^3 \ge 3x - 2$ so $x^3 + 2 y^2 + 3z \ge 3x - 2 + 3z + 2 y^2 \ge 2(x+y+z)$ using $ x , y , z \ge 1$ Doing the same thing we obtain $$4 y^3 + 5 z^2 + 6x \ge 5(x+y+z)$$and $$7 z^3 + 8 x^2 + 9y \ge 8(x+y+z) $$so $$ LHS \ge 80 (x+y+z)^3 \ge 720(xy+yz+zx) $$using again that $ x , y , z \ge 1$

Expanding the LHS we obtain \begin{align*} (x^3+2y^2+3z)(4y^3+5z^2+6x)(7z^3+8x^2+9y)&=(48x^6+72y^6+105z^6)\\&+(54x^4y+32x^5y^3+96x^3y^2+64x^2y^5+108xy^3+36x^3y^4)\\&+(42x^3z^3+40x^5z^2+144x^3z+35x^3z^5+126xz^4+120x^2z^3)\\&+(56y^5z^3+90y^3z^2+108y^4z+135z^3y+84z^4y^3+70y^2z^5)\\&+(28x^3y^3z^3+45x^3yz^2+84xy^2z^3+80x^2y^2z^2+162xyz+96x^2y^3z) \end{align*}Reducing all the exponents to $1$ we get a lower estimate of the expression in the second, third and fourth lines. To estimate the first line we use obvious inequality \begin{align*} 48x^6+72y^2+105z^6&=7.5(x^6+y^6)+64.5(y^6+z^6)+40.5(x^6+z^6)\\&\geq15x^3y^3+129y^3z^3+81x^3z^3\\&\geq15xy+129yz+81xz \end{align*}The expression in the first four lines is greater than or equal to $$405xy+588xz+672yz.$$It remains to check that the expression in the fifth line is not less than $$315xy+132xz+48yz,$$a task presenting no difficulties.

ACGNmath wrote: Prove the inequality $$(x^3+2y^2+3z)(4y^3+5z^2+6x)(7z^3+8x^2+9y)\geq720(xy+yz+xz)$$for $x, y, z \geq 1$. Proposed by K. Kokhas $$(x^3+2y^2+3z)(4y^3+5z^2+6x)(7z^3+8x^2+9y)\geq6\sqrt[6]{x^{3}y^{4}z^{3}}\cdot 15\sqrt[15]{x^{6}y^{12}z^{10}}\cdot 24\sqrt[24]{x^{16}y^{9}z^{21}}=3\cdot 720 x^{\frac{47}{30}}y^{\frac{221}{120}}z^{\frac{49}{24}}\geq 720(xy+yz+xz).$$