Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$

Question

$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$

A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,B,C$ are angles of an acute triangle.

Our inequality becomes $$\sqrt[2]{2(\cos A+ \cos B)}+\sqrt[3]{2(\cos B+\cos C)}+\sqrt[4]{2(\cos C+\cos A)} <4$$

I used formula $\cos(A)+\cos(B) < 2$ and estimate that $LHS <5$ but not $4$. I do not know how to proceed.

Answer 1

The condition $\;x^2+y^2+z^2+xyz=4\;$ can be rewritten as: $$ z^2+xy\cdot z + (x^2+y^2-4) = 0 \quad \Longrightarrow \quad z = - \frac{xy}{2} \pm \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)} $$ From $z \ge 0$ it follows that: $$ z = - \frac{xy}{2} + \sqrt{\left(\frac{xy}{2}\right)^2-(x^2+y^2-4)} \quad \mbox{and} \quad x^2+y^2\le 4 $$ Therefore, after substitution of $\,z$ , we only have to investigate function behavior
$f(x,y) = \sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} < 4$
in the area enclosed by x-axis, y-axis and a quarter of a circle, as depicted below.

Another proof without words is attempted by plotting a contour map of the function, as depicted. Levels (nivo) of these isolines are defined (in Delphi Pascal) as:

nivo := min + g/grens*(max-min); { grens = 40 ; g = 1..grens }

The blackness of the isolines is proportional to the (positive) function values; they are almost white near the minimum and almost black near the maximum values. Maximum and minimum values of the function are observed to be:

 2.54387743763872E+0000 < f < 3.91477606446737E+0000

The $\color{blue}{\mbox{blue}}$ spot is where $\,\left| f(x,y) - 4 \right| < 0.9$ . It is suggested by the rather large error $\,0.9\,$ that $\,4\,$ is not really the maximum. Indeed, upon refinement of the grid we find for the maximum numerically (double precision) a somewhat lower value:

3.91477205860402 < 4

Answer 2

Solution of 18.07.16

Using OP idea.

Let $$u=\dfrac{x}2,\quad v=\dfrac{y}2,$$ then translate the condition in form $$4u^2+4v^2+z^2+4uvz=4,$$ $$(z+2uv)^2 = 4(1-u^2)(1-v^2),$$ $$z=2\sqrt{(1-u^2)(1-v^2)}-2uv \leq (1-u^2)+(1-v^2)-2uv,$$ $$z\leq2-(u+v)^2.$$ So we can prove inequality $$\sqrt[2]{2(u+v)}+\sqrt[3]{2v+2-(u+v)^2}+\sqrt[4]{2u+2-(u+v)^2} < 4.$$ Using Cauchy-Schwarz inequality, we get $$\left(\sqrt[2]{2(u+v)}+\sqrt[3]{2v+2-(u+v)^2}+\sqrt[4]{2u+2-(u+v)^2}\right)^2$$ $$\leq (\sqrt[2]4+\sqrt[3]4+\sqrt[4]4)\left(u+v+\left(v+1-\dfrac12(u+v)^2\right)^{2/3}+\left(u+1-\dfrac12(u+v)^2\right)^{1/2}\right) < 16.$$ Maximal value of $$f(u,v) = u+v+\left(v+1-\dfrac12(u+v)^2\right)^{2/3}+\left(u+1-\dfrac12(u+v)^2\right)^{1/2}$$ can be achieved in local extremum within the area $u,v\in[0,1],$ or in the edges of the area.

The nesessary conditions of local extremum are $$f'_u=0,\quad f'_v=0,$$ or $$\begin{cases} 1+\dfrac23\dfrac{-u-v}{\sqrt[3]{v+1-\dfrac12(u+v)^2}}+\dfrac12\dfrac{1-u-v}{\sqrt{u+1-\dfrac12(u+v)^2}} = 0\\ 1+\dfrac23\dfrac{1-u-v}{\sqrt[3]{v+1-\dfrac12(u+v)^2}}+\dfrac12\dfrac{-u-v}{\sqrt{u+1-\dfrac12(u+v)^2}} = 0,\\ \end{cases}$$

$$\begin{pmatrix} -u-v & 1-u-v\\ 1-u-v & -u-v \end{pmatrix} \begin{pmatrix} \dfrac2{3\sqrt[3]{v+1-\dfrac12(u+v)^2}}\\ \dfrac1{2\sqrt{u+1-\dfrac12(u+v)^2}} \end{pmatrix} =\begin{pmatrix} -1\\-1 \end{pmatrix} $$ Solving this linear system by Cramer's rule, we obtain: $$ \Delta = \begin{vmatrix} -u-v & 1-u-v\\ 1-u-v & -u-v \end{vmatrix} = 2u+2v-1, $$ $$ \Delta_1 = \begin{vmatrix} -1 & 1-u-v\\ -1 & -u-v \end{vmatrix} = 1, $$ $$ \Delta_2 = \begin{vmatrix} -u-v & -1\\ 1-u-v & -1 \end{vmatrix} = 1, $$ $$\begin{cases} 3\sqrt[3]{v+1-\dfrac12(u+v)^2}= 2(2u+2v-1)\\ 2\sqrt{u+1-\dfrac12(u+v)^2}= 2u+2v-1 \end{cases}$$ If $$2(u+v)\geq1,$$ then $$\begin{cases} 27\left(2v+2-(u+v)^2\right)=16(2u+2v-1)^3\\ 2\left(2u+2-(u+v)^2\right)=(2u+2v-1)^2, \end{cases}$$ $$\begin{cases} 108\left(u+v+2-(u+v)^2\right)=32(2u+2v-1)^3+27(2u+2v-1)^2\qquad[2\times(1)+27\times(2)]\\ 2\left(2u+2-(u+v)^2\right)=(2u+2v-1)^2. \end{cases}$$ $$\begin{cases} 256(u+v)^3-168(u+v)^2-24(u+v)-221 = 0\\ u = \dfrac14(6(u+v)^2-4(u+v)-3) \end{cases}$$

Cubic equation $$256s^3-168s^2-24s-221 = 0$$ has one real root $$s = \dfrac1{32}\left(7+3^{5/3}\sqrt[3]{61-8\sqrt{58}}+3^{5/3}\sqrt[3]{61+8\sqrt{58}}\right),$$ so $f(u,v)$ has the unique extremum $$(u,v,f) \approx(0.392428, 0.875106, 3.08237).$$

The edges of the field achieves when $u\in\{0,1\}$ or $v\in\{0,1\}.$

$$f(0,v) = v + (v+1-\dfrac12v^2)^(2/3) + (1-\dfrac12v^2)^(1/2)$$ has the global maximum $\approx3.03678,$ so $$f(0,v) < 3.08237.$$

$$f(1,v) = v+1 + (v+1-\dfrac12(1+v)^2)^(2/3) + (2-\dfrac12(1+v)^2)^(1/2)$$ has the global maximum $\approx 2.9714,$ so $$f(1,v) < 3.08237.$$

$$f(u,0) = u + (1-\dfrac12u^2)^(2/3) + (1+u-\dfrac12u^2)^(1/2)$$ has the global maximum $\approx 2.86142,$ so $$f(u,0) < 3.08237.$$

$$f(u,1) = u+1 + (2-\dfrac12(u+1)^2)^(2/3) + (u+1-\dfrac12(u+1)^2)^(1/2)$$ has the global maximum $\approx 3.07604,$ so $$f(u,1) < 3.08237.$$ Note that $$f(0,0)=2,\quad f(0,1)\approx 3.017477,\quad f(1,0)\approx 2.854705, \quad f(1,1)=2.$$

Given the conditions and the accuracy of the calculations this means that $$f(u,v) < 3.08238,$$ $$\left(\sqrt[2]{2(u+v)}+\sqrt[3]{2v+2-(u+v)^2}+\sqrt[4]{2u+2-(u+v)^2}\right)^2 < 3.08238\left(\sqrt[2]4+\sqrt[3]4+\sqrt[4]4\right) < 15.4169,$$ $$\sqrt[2]{2(u+v)}+\sqrt[3]{2v+2-(u+v)^2}+\sqrt[4]{2u+2-(u+v)^2} < 3.926,$$ and for $\quad x,y,z \geq 0,\quad x^2+y^2+z^2+xyz=4$ $$\boxed{\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}<4.}$$

Answer 3

Remark: My second proof is given in https://artofproblemsolving.com/community/c6h2483725p20868007.
There, KaiRain@AoPS gave a nice proof.

Proof without using derivative

We split into three cases:

1. $y+z \le 1$ and $z+x \le 1$: We have $x + y \le 2$. Thus, we have $\mathrm{LHS} \le \sqrt{2} + 1 + 1 < 4$. The inequality is true.

2. $y + z > 1$ and $z + x \le 1$: Since $\sqrt[3]{y+z} < \sqrt{y+z}$ and $\sqrt[4]{z+x} \le 1$, it suffices to prove that $$\sqrt{x+y} + \sqrt{y+z} \le 3.$$ By AM-QM inequality, we have $\sqrt{x+y} + \sqrt{y+z} \le \sqrt{2(x+y + y + z)}$. Thus, It suffices to prove that $$x+2y+z \le \frac{9}{2}.$$ From $x^2+y^2+z^2 + xyz = 4$, we have $y^2 + (\frac{1}{2} + \frac{y}{4})(x + z)^2 + (\frac{1}{2} - \frac{y}{4})(x - z)^2 = 4$ which results in $y^2 + (\frac{1}{2} + \frac{y}{4})(x + z)^2 \le 4$ and $x + z \le \sqrt{\frac{4(4 - y^2)}{2 + y}}$. Thus, it suffices to prove that $2y + \sqrt{\frac{4(4 - y^2)}{2 + y}} \le \frac{9}{2}$. It suffices to prove that $(\frac{9}{2} - 2y)^2 \ge \frac{4(4 - y^2)}{2 + y}$ that is $\frac{1}{4}(4y - 7)^2 \ge 0$. The inequality is true.

3. $z + x > 1$: Since $\sqrt[3]{z+x} > \sqrt[4]{z+x}$, it suffices to prove that $$ \sqrt{x+y} + \sqrt[3]{y+z} + \sqrt[3]{z+x} < 4. $$ Note that $u\mapsto \sqrt[3]{u}$ is concave on $u > 0$. It suffices to prove that $$ \sqrt{x+y} + 2\sqrt[3]{\frac{y+z + z+x}{2}} < 4. $$ From $x^2+y^2+z^2 + xyz = 4$, we have $(z + \frac{1}{2}xy)^2 = \frac{1}{4}(4-x^2)(4-y^2)$ which results in $$z = \frac{1}{2}\sqrt{(4-x^2)(4-y^2)}-\frac{1}{2}xy \le \frac{1}{2}\frac{4-x^2+4-y^2}{2} - \frac{1}{2}xy = 2- \frac{(x+y)^2}{4}.$$ Thus, it suffices to prove that $$ \sqrt{x+y} + 2\sqrt[3]{\frac{y+x}{2} + 2- \frac{(x+y)^2}{4}} < 4. $$ Let $v = \sqrt{x+y}$. Since $x+y \le \sqrt{2(x^2+y^2)} \le \sqrt{2 \cdot 4} < 4$, we have $v\in (0, 2)$. It suffices to prove that $$v + 2\sqrt[3]{\frac{v^2}{2} + 2- \frac{v^4}{4}} < 4$$ or $$\frac{v^2}{2} + 2- \frac{v^4}{4} < \frac{1}{8}(4 - v)^3$$ or $$\frac{1}{8}(2v^4-v^3+8v^2-48v+48) > 0$$ for $v\in (0, 2)$. It is easy to prove that $2v^4-v^3+8v^2-48v+48 > 0$ for $v\in (0, 2)$. We are done.

Answer 4

PARTIAL SOLUTION

$$f(x,y,z)=\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}$$ $$g(x,y,z)= x^2+y^2+z^2+xyz-4=0$$ $$ \begin{align}\mathcal{L}(x,y,z,\lambda) =& \;f(x,y,z) + \lambda g(x,y,z)\\ =&\;\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}+\lambda(x^2+y^2+z^2+xyz-4) \end{align}$$ $$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambda) =\left(\frac{\partial \mathcal{L}}{\partial x},\frac{\partial \mathcal{L}}{\partial y},\frac{\partial \mathcal{L}}{\partial z},\frac{\partial \mathcal{L}}{\partial \lambda}\right)$$ Now comes the work finding the partial derivatives

$$\begin{align} &\frac{\partial \mathcal{L}}{\partial x} =\; \lambda(2x+yz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{4(x+z)^{3/4}}\\ &\frac{\partial \mathcal{L}}{\partial y} =\; \lambda(2y+xz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{3(y+z)^{2/3}}\\ &\frac{\partial \mathcal{L}}{\partial z} =\; \lambda(2z+xy) + \frac{1}{3(y+z)^{2/3}}+\frac{1}{4(x+z)^{3/4}}\\ &\frac{\partial \mathcal{L}}{\partial \lambda} =\; x^2+y^2+z^2+xyz-4 \end{align}$$

$$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambda)=0 \iff \begin{cases} \lambda(2x+yz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{4(x+z)^{3/4}} & = 0 \\ \lambda(2y+xz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{3(y+z)^{2/3}} & = 0 \\ \lambda(2z+xy) + \frac{1}{3(y+z)^{2/3}}+\frac{1}{4(x+z)^{3/4}} & = 0 \\ x^2+y^2+z^2+xyz-4, & = 0 \end{cases}$$ At this point I cannot seem to solve the resulting equations. The symmetry involved implies some simple trick should suffice, though I have yet to spot one.