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Let $x,y,z>0$ such that $x+y+z=3$. Prove that $$\sum x\sqrt{x^3+3y} \ge 6$$

This trying doesn't help. With Cauchy Schwarz

$(\sum x\sqrt{x^3+3y})^2\geq \sum x^2\sum(x^3+3y) = (x^2 + y^2 + z^2)(x^3+y^3+z^3+9) \geq 3(xyz)^{2/3}(3xyz + 9)$

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

Hence $\sum x\sqrt{x^3+3y}\geq 6.$

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  • $\begingroup$ Should there be a $z$ in the inequality? $\endgroup$ – Jack M Aug 23 '18 at 5:53
  • $\begingroup$ Please try to choose titles with strictly more than zero content. $\endgroup$ – Did Aug 23 '18 at 6:01
  • $\begingroup$ What does the sum mean? $\endgroup$ – 4-ier Aug 23 '18 at 6:30
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    $\begingroup$ @4-ier The inequality is $\sum_{cyc} x\sqrt{x^3+3y}\ge 6.$ In other words $\sum x\sqrt{x^3+3y}+\sum y\sqrt{y^3+3z}+\sum z\sqrt{z^3+3x} \ge 6.$ $\endgroup$ – mfl Aug 23 '18 at 6:56
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The hint.

By Minkowski $$\sum_{cyc}x\sqrt{x^3+3y}=\sum_{cyc}\sqrt{x^5+3x^2y}\geq\sqrt{\left(\sum\limits_{cyc}\sqrt{x^5}\right)^2+3\left(\sum\limits_{cyc}\sqrt{x^2y}\right)^2}.$$ Thus, it's enough to prove that $$\left(\sqrt{x^5}+\sqrt{y^5}+\sqrt{z^5}\right)^2+3\left(\sqrt{x^2y}+\sqrt{y^2z}+\sqrt{z^2x}\right)^2\geq36,$$ which is true.

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

Now, by AM-GM and Rearrangement we obtain: $$(x^2y+y^2z+z^2x)^2=\sum_{cyc}(x^4y^2+2x^3z^2y)\geq$$ $$\geq7x^2y^2z^2+\sum_{cyc}(x^4y^2+x^4z^2)-(x^4z^2+y^4x^2+z^4y^2+x^2y^2z^2)\geq$$ $$\geq7x^2y^2z^2+\sum_{cyc}(x^4y^2+x^4z^2)-\frac{4}{27}(x^2+y^2+z^2)^3=$$ $$=4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-\frac{4}{27}(x^2+y^2+z^2)^3.$$ Thus, it's enough to prove that $$(x^5+y^5+z^5)^2+$$ $$+3\left(4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-\frac{4}{27}(x^2+y^2+z^2)^3\right)\geq36$$ or $$3(x^5+y^5+z^5)^2+$$ $$+(x^2+y^2+z^2)^2\left(4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)\right)\geq\frac{16}{27}(x^2+y^2+z^2)^5.$$ 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 $$3(x^5+y^5+z^5)^2+$$ $$+(x^2+y^2+z^2)^2\left(4x^2y^2z^2+(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)\right)\geq\frac{16}{27}(x^2+y^2+z^2)^5$$ for the minimal value of $w^3$, which happens in the following cases.

  1. $w^3\rightarrow0^+$.

Let $z\rightarrow0^+$ and $y=1$. We can assume it because the last inequality is homogeneous.

Thus, we need to prove that $$3(x^5+1)^2+(x^2+1)^3x^2\geq\frac{16}{27}(x^2+1)^5.$$ Let $x^2+1=2tx$.

Hence, by AM-GM $t\geq1$ and we need to prove that $$3(x+1)^2(x^4-x^3+x^2-x+1)^2+(x^2+1)^3x^2\geq\frac{16}{27}(x^2+1)^5$$ or $$3(t+1)(4t^2-2t-1)^2+4t^3\geq\frac{256}{27}t^5$$ or $$1040t^5-1512t^3+405t+81\geq0$$ or $$(t-1)(1040t^4+1040t^3-472t^2-472t-67)+14>0,$$ which is obvious for $t\geq1$.

  1. Two variables are equal.

Let $y=z=1$.

Thus, we need to prove that $$3(x^5+2)^2+(x^2+2)^2\left(4x^2+(x^2+2)(2x^2+1)\right)\geq\frac{16}{27}(x^2+2)^5$$ 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!

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    $\begingroup$ I'm new to Minkowski, can you clarify your use here? Looking at Wolfram & Wiki isn't the p supposed to be greater than 1? $\endgroup$ – Andrew Allen Aug 23 '18 at 7:48
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    $\begingroup$ @Andrew Allen Minkowski it's the following $\sqrt{a^2+x^2}+\sqrt{b^2+y^2}+\sqrt{c^2+z^2}\geq\sqrt{(a+b+c)^2+(x+y+z)^2}$. It's just the triangle inequality. $\endgroup$ – Michael Rozenberg Aug 23 '18 at 7:50
  • $\begingroup$ I think it should be $\sqrt{3}$ $\endgroup$ – Truth Aug 23 '18 at 8:14
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    $\begingroup$ Unruly Kid Are you sure? I think $(\sqrt3)^2=3.$ $\endgroup$ – Michael Rozenberg Aug 23 '18 at 8:16
  • $\begingroup$ Yes, it is 3 and how to prove the last inequality $\endgroup$ – Truth Aug 23 '18 at 8:17

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