[tex]\bf m_a\geq m_g\\ \\ \it\begin{cases} \it x+2\geq2\sqrt{2x}\\ \\ \it 2y+3\geq2\sqrt{6y}\\ \\ \it 3z+4\geq2\sqrt{12z}\end{cases}\ \ \Rightarrow (x+2)(2y+3)(3z+4)\geq8\sqrt{144xyz}=96\sqrt{xyz}[/tex]

Verified answer

Inegalitatea mediilor:

[tex]\boldsymbol{m_{g} \leq m{a} \iff \sqrt{xy} \leqslant \dfrac{x + y}{2}}, \ x,y>0 \\ [/tex]

[tex]\dfrac{x + 2}{2} \geqslant \sqrt{2x} \iff x + 2 \geqslant 2 \sqrt{2x} [/tex]

[tex]\dfrac{2y + 3}{2} \geqslant \sqrt{2y \times 3} \iff 2y + 3 \geqslant 2 \sqrt{6y} \\ [/tex]

[tex]\dfrac{3z + 4}{2} \geqslant \sqrt{3z \times 4} \iff 3z + 4 \geqslant 2 \sqrt{12z} \\ [/tex]

Înmulțim:

[tex](x+2)(2y+3)(3z+4) \geqslant 2 \sqrt{2x} \times 2 \sqrt{6y} \times 2 \sqrt{12z} \\ [/tex]

[tex](x+2)(2y+3)(3z+4) \geqslant 8 \sqrt{2x \times 6y \times 12z} \\ [/tex]

[tex](x+2)(2y+3)(3z+4) \geqslant 8 \sqrt{ {12}^{2}xyz} \\ [/tex]

[tex](x+2)(2y+3)(3z+4) \geqslant 8 \times 12 \sqrt{xyz} \\ [/tex]

[tex]\implies \boldsymbol{ (x+2)(2y+3)(3z+4) \geqslant 96 \sqrt{xyz}} \\ [/tex]

q.e.d.