inequality

[Victoria124]

How I prove this inequality : If x,y,z =>2, then (y^3+x)(z^3+y)(x^3+z)=>125xyz ?

 

[stapel]

How I prove this inequality : If x,y,z =>2, then (y^3+x)(z^3+y)(x^3+z)=>125xyz ?

Online solutions are available, but are somewhat cryptic, at least to me. Start with this:

. . . . .\(\displaystyle f(x)\, =\, x^3\, -\, 4x\, \geq\, 0\, \mbox{ for }\, x\, \geq\, 2\)

(If you're not sure of this inequality, solve for the x-intercepts of the function, and check the graph of the cubic.) This says that:

. . . . .\(\displaystyle x^3\, \geq\, 4x\, \mbox{ for }\, x\, \geq\, 2\)

...and similarly for \(\displaystyle y\) and \(\displaystyle z\). The above leads to:

. . . . .\(\displaystyle y^3\, +\, x\, \geq\, 4y\, +\, x\, =\, (y\, +\, y\, +\, y\, +\, y)\, +\, x\)

Then some "magic" happens. Somehow (?!?) one is expected to consider the following:

. . . . .\(\displaystyle \sqrt[5]{y^4 x\, }\)

Suppose \(\displaystyle x\, \leq\, y\). Then:

. . . . .\(\displaystyle \sqrt[5]{y^4 x\, }\, \leq\, \sqrt[5]{y^5\, }\, =\, y\)

On the other hand, suppose \(\displaystyle x\, >\, y\). Then:

. . . . .\(\displaystyle \sqrt[5]{y^4 x\, }\, <\, \sqrt[5]{x^5\, }\, =\, x\)

Either way, you get:

. . . . .\(\displaystyle (y\, +\, y\, +\, y\, +\, y)\, +\, x\, \geq\, \left(\sqrt[5]{y^4 x\, }\, +\, \sqrt[5]{y^4 x\, }\, +\, \sqrt[5]{y^4 x\, }\, +\, \sqrt[5]{y^4 x\, }\right)\, +\, \sqrt[5]{y^4 x\, }\, =\, 5 \sqrt[5]{y^4 x\, }\)

See where that leads.

 

[Victoria124]

I didn' learn at highschool what did you explain there.I really appreciate your effort but I don't understand

 

[Victoria124]

The last time he tought us the chauchy-bunyakovski-schwatrz inequality and Mikovski inequality