0
$\begingroup$

Prove that forl all positive real numbers $x, y, z$ we have that $ (x^3 + y^3 + z^3)^2 \geq 3(x^2y^4 + y^2z^4 + z^2x^4)$. I tried to apply Cebasev, Muirhead but doesn't work.

$\endgroup$
1
$\begingroup$

\begin{eqnarray*} y^3(2x+y)(x-y)^2+z^3(2y+z)(y-z)^2+x^3(2z+x)(z-x)^2 \geq 0 \end{eqnarray*} Now rearrange.

$\endgroup$
0
$\begingroup$

Because $$(x^3+y^3+z^3)^2-3(x^4z^2+y^4x^2+z^4y^2)=$$ $$=\sum_{cyc}(x^6+2x^3y^3-3x^2y^4)=\sum_{cyc}x^2(x^4+2xy^3-3y^4)=$$ $$=\sum_{cyc}x^2(x^4-x^3y+x^3y-x^2y^2+x^2y^2-xy^3+3xy^3-3y^4)=$$ $$=\sum_{cyc}x^2(x-y)(x^3+x^2y+xy^2+3y^3)=$$ $$=\sum_{cyc}\left((x-y)(x^5+x^4y+x^3y^2+3x^2y^3-(x^6-y^6)\right)=$$ $$=\sum_{cyc}(x-y)(2x^2y^3-xy^4-y^5)=\sum_{cyc}y^3(x-y)^2(2x+y)\geq0.$$ Also, there is a proof by Rearrangement and AM-GM, but it's much more complicated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.