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Let $x, y, z$ be positive real numbers such that $x + y + z = xyz$.

Find the minimum value of the expression $$\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$$

My attempt: By using Cauchy-Schwarz inequality and AM-GM inequality, $$\sqrt{\frac{1}{3}t^4+1}\cdot\sqrt{3+1}\geq t^2+1$$ $$(x^2+y^2+z^2)(x+y+z)\geq3(xyz)^{\frac{2}{3}}\cdot 3(xyz)^{\frac{1}{3}}=9xyz$$ $$\therefore \sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}\geq \frac{1}{2}(x^2+y^2+z^2)+\frac{3}{2}$$ $$\geq\frac{1}{2}\left (\frac{9xyz}{x+y+z}+\frac{3}{2} \right)=6$$ with equality if and only if $x=y=z=\sqrt{3}$

What are the alternative methods to solve this question?

Mathxx
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  • For what I see, it seems that the method you propose just gives an inequality, not the minimum. Maybe you should explain more in detail. Moreover: is there any reason why the Lagrange multiplier method cannot be applied? – Doriano Brogioli Nov 08 '22 at 16:44
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    @DorianoBrogioli The answer proposed is complete, as the inequality is not just a lower bound, $6$ can be achieved by the choice of $x=y=z=\sqrt3$, hence that's indeed the global minimum under the constraints given. In fact while alternate methods including Lagrange multipliers exist, arguably they'll take more effort in finding and establishing the minimum. – Macavity Nov 08 '22 at 17:29
  • @Macavity: I have put at denominator $0.1^2+1$ instead of $0.1^2-1$. Thanks you. – Piquito Nov 09 '22 at 21:31

2 Answers2

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You already have a good method, here is an alternative with only AM-GM. First, we have $xyz = x+y+z \geqslant 3\sqrt[3]{xyz}\implies xyz \geqslant 3\sqrt3$. Further, again using AM-GM,

$$\sqrt{\frac{t^4}9+\frac{t^4}9+\frac{t^4}9+1}\geqslant \sqrt{\frac4{3\sqrt3}t^3}=\frac2{3^{3/4}}t^{3/2}$$

$\begin{align} \implies \sqrt{\frac13x^4+1}+\sqrt{\frac13y^4+1}+\sqrt{\frac13z^4+1} &\geqslant \frac2{3^{3/4}}(x^{3/2}+y^{3/2}+z^{3/2}) \\ &\geqslant \frac2{3^{3/4}}\cdot 3\,(xyz)^{1/2} \\ &\geqslant \frac6{3^{3/4}}\,(3\sqrt3)^{1/2}=6 \end{align} $

with equality possible iff $x=y=z=\sqrt3$, so we have the minimum.

P.S. Yet another way would be to show the estimate $\sqrt{t^4/3+1} \geqslant \sqrt3 t -1$ holds, and then $\sum \sqrt{x^4/3+1} \geqslant \sqrt3(\sum x)-3 = \sqrt3(xyz)-3 \geqslant \sqrt3\cdot 3\sqrt3-3 = 6$

Macavity
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2

Your method is very straightforward. I am going just to post another approach that may contain a new idea.

Firstly; $$xyz=x+y+z\ge 3 \sqrt[3] {xyz}\implies (xyz)^3\ge 27xyz\implies x+y+z=xyz\ge 3\sqrt3.$$

Secondly, a simple computation shows that $f(t)=\sqrt{\frac {t^4}{3}+1}$ is a convex function. Therefore, by Jensen's inequality, we get:

$$\sqrt{\frac {x^4}{3}+1}+\sqrt{\frac {y^4}{3}+1}+\sqrt{\frac {z^4}{3}+1}\ge 3\sqrt {\frac {(\frac {x+y+z}{3})^4}{3}+1}\ge 3\sqrt {\frac {(\frac {3\sqrt 3}{3})^4}{3}+1}=6. $$

More information about Jensen's inequality is available here.

Reza Rajaei
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