Vtyjkyuk

Let $x,y,z>0$ such that $x+y+z=3$. Prove that
$$sum xsqrtx^3+3y ge 6$$

This trying doesn't help.
With Cauchy Schwarz

$(sum xsqrtx^3+3y)^2geq sum x^2sum(x^3+3y) = (x^2 + y^2 + z^2)(x^3+y^3+z^3+9) geq 3(xyz)^2/3(3xyz + 9)$

Then the maximum value of $xyz$ is $1$.

Hence $sum xsqrtx^3+3ygeq 6.$

The hint.

By Minkowski
$$sum_cycxsqrtx^3+3y=sum_cycsqrtx^5+3x^2ygeqsqrtleft(sumlimits_cycsqrtx^5right)^2+3left(sumlimits_cycsqrtx^2yright)^2.$$
Thus, it's enough to prove that
$$left(sqrtx^5+sqrty^5+sqrtz^5right)^2+3left(sqrtx^2y+sqrty^2z+sqrtz^2xright)^2geq36,$$ which is true.

Indeed, after replacing $x$ by $x^2$, $y$ by $y^2$ and $z$ by $z^2$ we need to prove that
$$(x^5+y^5+z^5)^2+3(x^2y+y^2z+z^2x)^2geq36$$
for positives $x$, $y$ and $z$ such that $x^2+y^2+z^2=3.$

Now, by AM-GM and Rearrangement we obtain:
$$(x^2y+y^2z+z^2x)^2=sum_cyc(x^4y^2+2x^3z^2y)geq$$
$$geq7x^2y^2z^2+sum_cyc(x^4y^2+x^4z^2)-(x^4z^2+y^4x^2+z^4y^2+x^2y^2z^2)geq$$
$$geq7x^2y^2z^2+sum_cyc(x^4y^2+x^4z^2)-frac427(x^2+y^2+z^2)^3=$$
$$=4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-frac427(x^2+y^2+z^2)^3.$$
Thus, it's enough to prove that
$$(x^5+y^5+z^5)^2+$$
$$+3left(4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-frac427(x^2+y^2+z^2)^3right)geq36$$ or
$$3(x^5+y^5+z^5)^2+$$
$$+(x^2+y^2+z^2)^2left(4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)right)geqfrac1627(x^2+y^2+z^2)^5.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $f(w^3)geq0,$ where
$$f(w^3)=3(81u^5-135u^3v^2+45uv^4+15u^2w^3-5v^2w^3)^2+$$
$$+(3u^2-2v^2)^2(4w^6+(9u^2-6v^2)(9v^4-6uw^3))-16(3u^2-2v^2)^5.$$
But by Schur we obtain:
$$f'(w^3)=6(81u^5-135u^3v^2+45uv^4+15u^2w^3-5v^2w^3)(15u^2-5v^2)+$$
$$+8(3u^2-2v^2)^2w^3-6u(3u^2-2v^2)^2(9u^2-6v^2)=$$
$$=2(3402u^7-6804u^5v^2+3726u^3v^4-603uv^6+(711u^4-498u^2v^2+91v^4)w^3)geq$$
$$geq2(3402u^7-6804u^5v^2+3726u^3v^4-603uv^6+(711u^4-498u^2v^2+91v^4)(4uv^2-3u^3))=$$
$$=2u(1269u^6-2466u^4v^2+1461u^2v^4-239v^6)=$$
$$=2u(1269u^6-1269u^4v^2-1197u^4v^2+1197u^2v^4+264u^2v^4-264v^6+25v^6)=$$
$$=2u((u^2-v^2)(1269u^4-1197u^2v^2+264v^4)+25v^6)>0,$$ which says that $f$ increases.

Thus, it's enough to prove that
$$3(x^5+y^5+z^5)^2+$$
$$+(x^2+y^2+z^2)^2left(4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)right)geqfrac1627(x^2+y^2+z^2)^5$$ for the minimal value of $w^3$, which happens in the following cases.

1. $w^3rightarrow0^+$.

Let $zrightarrow0^+$ and $y=1$. We can assume it because the last inequality is homogeneous.

Thus, we need to prove that
$$3(x^5+1)^2+(x^2+1)^3x^2geqfrac1627(x^2+1)^5.$$
Let $x^2+1=2tx$.

Hence, by AM-GM $tgeq1$ and we need to prove that
$$3(x+1)^2(x^4-x^3+x^2-x+1)^2+(x^2+1)^3x^2geqfrac1627(x^2+1)^5$$ or
$$3(t+1)(4t^2-2t-1)^2+4t^3geqfrac25627t^5$$ or
$$1040t^5-1512t^3+405t+81geq0$$ or
$$(t-1)(1040t^4+1040t^3-472t^2-472t-67)+14>0,$$ which is obvious for $tgeq1$.

2. Two variables are equal.

Let $y=z=1$.

Thus, we need to prove that
$$3(x^5+2)^2+(x^2+2)^2left(4x^2+(x^2+2)(2x^2+1)right)geqfrac1627(x^2+2)^5$$ or
$$(x-1)^2(65x^8+130x^7+89x^6+48x^5-174x^4-72x^3-8x^2+56x+28)geq0,$$
which is true by AM-GM.

Done!