If $x+y+z=1$ Find the maximum of $\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}$

Question

Question -

Let $x, y, z$ be non-negative real numbers such that $x+y+z=1 .$ Find the maximum of $$ \frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}} $$

My work -

first WLOG $x\ge y\ge z$

I rewrite given inequality like this $\frac{x-y}{x+y}{\sqrt{x+y}}+\frac{y-z}{y+z}{\sqrt{y+z}}+\frac{z-x}{z+x}{\sqrt{z+x}}$

and now applying C-S and using fact that $(\frac{x-y}{x+y})^2 < 1 $ and $x+y+z=1$

i get maximum value as ${\sqrt6}$.

Is my proof correct ???

thankyou

Answer 1

Your answer is not correct because your value $\sqrt6$ does not occur.

Also, you need to prove that we can assume $x\geq y\geq z$.

My solution:

Let $z=\min\{x,y,z\}$.

Now, we'll prove that if $y\geq x\geq z$ so $$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq\frac{y-x}{\sqrt{x+y}}+\frac{x-z}{\sqrt{x+z}}+\frac{z-y}{\sqrt{z+y}},$$ which will say us that we can assume that $x\geq y\geq z$.

Indeed, we need to prove that $$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq0$$ or $$\frac{x-y}{\sqrt{x+y}}+\frac{y-x+x-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}\leq0$$ or $$(y-x)\left(\frac{1}{\sqrt{y+z}}-\frac{1}{\sqrt{x+y}}\right)\leq(x-z)\left(\frac{1}{\sqrt{x+z}}-\frac{1}{\sqrt{y+z}}\right)$$ or $$\frac{(y-x)(x-z)}{\sqrt{(x+y)(y+z)}(\sqrt{x+y}+\sqrt{y+z})}\leq\frac{(y-x)(x-z)}{\sqrt{(x+z)(y+z)}(\sqrt{x+z}+\sqrt{y+z})}$$ or $$x+z+\sqrt{(x+z)(y+z)}\leq x+y+\sqrt{(x+y)(y+z)},$$ which is obvious.

Now, let $$f(z)=\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}},$$ where $x\geq y\geq z$.

Thus, $$f'(z)=\frac{3x+z}{2\sqrt{(x+z)^3}}-\frac{3y+z}{2\sqrt{(y+z)^3}}=$$ $$=\frac{(3x+z)^2(y+z)^3-(3y+z)^2(x+z)^3}{2\sqrt{(x+z)^3(y+z)^3}((3x+z)\sqrt{(y+z)^3}+(3y+z)\sqrt{(x+z)^3})}=$$ $$=\frac{(y-x)(9x^2y^2+6xyz(x+y)+(x^2+y^2-8xy)z^2-6(x+y)z^3-3z^4)}{2\sqrt{(x+z)^3(y+z)^3}((3x+z)\sqrt{(y+z)^3}+(3y+z)\sqrt{(x+z)^3})}\leq0.$$ Id est, $f$ decreases, which says that it's enough to assume $z=0$ and we obtain: $$\max f=\max f(0)=\max_{x+y=1,x\geq y>0}\left(x-y+\sqrt{y}-\sqrt{x}\right)=$$ $$=\max_{\frac{1}{2}\leq x\leq 1}(2x-1+\sqrt{1-x}-\sqrt{x})=\sqrt{\frac{71-17\sqrt{17}}{32}},$$ where $$x_{max}=\frac{1}{2}+\sqrt{\frac{23-\sqrt{17}}{128}}.$$ Can you get this answer?

It's interesting, that the equation, which we obtain, when the derivative is equal to $0$, is nice enough: $$4\sqrt{x(1-x)}=\sqrt{1-x}+\sqrt{x},$$ which we can solve by the substitution $$\sqrt{1-x}+\sqrt{x}=t.$$