# Find the Maximum value of $\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$

if $x$, $y$ and $z$ are positive real numbers such that $x+y+z=4$ Find the maximum value of $$S=\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}$$

I tried as follows.

The given expression can be rewritten as

$$S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}-\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$

But by symmetry $$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$

so

$$2S=\sqrt{4-x}+\sqrt{4-y}+\sqrt{4-z}$$ and by Cauchy Scwartz inequality

$$2S \le \sqrt{4-x+4-y+4-z}\times \sqrt{3}$$ so

$$2S \le \sqrt{24}$$

so

$$S \le \sqrt{6}$$

Is this approach correct?

• There's the approach of substituting $z=4-x-y$ and then find the maximum of the resulting surface. BTW if you set $x=y=z$, then $S=\sqrt{6}$. – Χpẘ Apr 24 '17 at 18:19
• $S=\sqrt6$ is maximum. $S(3,1,0)=\dfrac{3+\sqrt3}2<\sqrt6.$ – Yuri Negometyanov Jul 10 '18 at 7:14

I think it means that $x$, $y$ and $z$ are non-negatives such that $xy+xz+yz\neq0$.

If $x=3$, $y=1$ and $z=0$ then $S=\frac{5}{2}$.

We'll prove that it's a maximal value.

Indeed, we need to prove that: $$\sum_{cyc}\frac {x}{\sqrt {x+y}}\leq\frac{5}{4}\sqrt{x+y+z}.$$

By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$ Id est, it remains to prove that $$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or $$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$ $$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or $$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$ $$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.

Done!

• How did you come up with $3,1,0$? – Χpẘ Apr 24 '17 at 19:00
• @Χpẘ It's just trying only, but it helped. – Michael Rozenberg Apr 24 '17 at 19:02
• @Χpẘ I have a version=) – Yuri Negometyanov Apr 27 '17 at 12:30
• @Χpẘ $S(3,1,0)=\dfrac{3+\sqrt3}2<\sqrt6.$ – Yuri Negometyanov Jul 10 '18 at 7:11

No, this approach is not correct, since

$$S=\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right)$$

does not necessarily hold. For example, consider $x=\frac12$, $y = \frac32$ and $z=2$, then $S=\left(\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}} \right) \approx 2.420$, but $\left(\frac{y}{\sqrt{x+y}}+\frac{z}{\sqrt{y+z}}+\frac{x}{\sqrt{z+x}}\right) \approx 2.445$.

Also, if this were correct, you would need to provide $x, y$ and $z$ such that $S = \sqrt{6}$.

• Ok so can i have any hint after my first step – Umesh shankar Apr 24 '17 at 18:03
• @Umeshshankar I'm sorry, I haven't found the solution of this problem myself yet. – wythagoras Apr 24 '17 at 18:28

$$\mathbf{\color{green}{New\ version\ of\ 10.02.2018}}$$

Let us find the greatest value of $\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$ under the condition $x+y+z=4.$

Substitutions $$a=\sqrt{x+y},\quad b=\sqrt{y+z},\quad c=\sqrt{z+x}$$ lead to the task on the greatest value of $a+b+c$ under the condition $a^2+b^2+c^2=8.$ Stationary points can be found using Lagrange multipliers method with the function $$f(a,b,c,\lambda)=a+b+c+\lambda(a^2+b^2+c^2-8)$$ via solution of the system $f'_a=f'_b=f'_c=f'_\lambda=0,$ or \begin{cases} 1+2\lambda a = 0\\ 1+2\lambda b = 0\\ 1+2\lambda c=0\\ a^2+b^2+c^2=8\\ (a,b,c)\in \mathbb R_+. \end{cases} Summation of $(1.1)-(1.4)$ with factors $a,b,c,-\lambda$ gives $$a+b+c=-16\lambda,$$ and after substitution of this to $(1)$ one can get the system $$a(a+b+c)=b(a+b+c)=c(a+b+c),\quad a^2+b^2+c^2=8,$$ with the evident solution $$a=b=c=\sqrt{\dfrac83},\quad x=y=z=\sqrt{\dfrac43},\quad f=\sqrt{24},$$ so $$\boxed{S=\sqrt6\approx2.449490}$$ is the least value. For example, in the point $(x,y,z)=(0,1,3)$ $$S_1=\dfrac{\sqrt1+\sqrt4+\sqrt3}2=\dfrac{3+\sqrt3}2\approx2.366025 < S.$$

• Shouldn't the maximum value be at $(3,1,0)$? $S(3,1,0) = \frac{5}{2}$. – Toby Mak Nov 3 '19 at 12:47