Find the maximum value of the expression : $${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$ where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
We use the method of Lagrange Multipliers.
First, we define the function to maximize as $f(x,y,z)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ and our constraint as $g(x,y,z)=x+y+z-1$.
Thus, the Lagrange function is:
$$\mathcal{L}(x,y,z,\lambda)=f(x,y,z)+\lambda\cdot g(x,y,z)$$
$$\mathcal{L}(x,y,z,\lambda)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}+\lambda(x+y+z-1) \tag{1}$$
Where $\lambda$ is some scalar. This function's partial derivatives must equal all equal to zero:
$$\frac{\partial \mathcal{L}}{\partial x}=\frac{1-x^2}{(x^2+1)^2}+\lambda=0$$ $$\frac{\partial \mathcal{L}}{\partial y}=\frac{1-y^2}{(y^2+1)^2}+\lambda=0$$ $$\frac{\partial \mathcal{L}}{\partial z}=\frac{1-z^2}{(z^2+1)^2}+\lambda=0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda}=x+y+z-1=0$$
Solving this system of equations will yield values for $x,y,z,\lambda$ which will either minimize or maximize $f(x,y,z)$.
After some tedious calculation, we notice that the solution which maximizes the value of $f(x,y,z)$ is:
$x=y=z=\frac{1}{3}$ and $\lambda=-0.72$, as you have correctly predicted.
Thus, the maximum value of $f(x,y,z)=\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}=\frac{9}{10}$.
Let $x=\frac{a}{3}$, $y=\frac{b}{3}$ and $z=\frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$\sum_{cyc}\frac{a}{9+a^2}\leq\frac{3}{10}$$ or $$\sum_{cyc}\left(\frac{1}{10}-\frac{a}{9+a^2}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2-10a+9}{9+a^2}\geq0$$ or $$\sum_{cyc}\left(\frac{a^2-10a+9}{9+a^2}+\frac{2}{3}\right)\geq2$$ or $$\sum_{cyc}\frac{(a-3)^2}{9+a^2}\geq\frac{6}{5}.$$ Let $k=3\sqrt2-6$. Hence, by C-S $$\sum_{cyc}\frac{(a-3)^2}{9+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(a+k)^2(a^2+9)}=\frac{\left(\sum\limits_{cyc}(a^2-3-2k)\right)^2}{\sum\limits_{cyc}(a+k)^2(a^2+9)}.$$ Thus, it remains to prove that $$5\left(\sum\limits_{cyc}(a^2-3-2k)\right)^2\geq6\sum\limits_{cyc}(a+k)^2(a^2+9),$$ which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain $$(a-1)^2(\sqrt2a-3\sqrt2+3)^2\geq0.$$ Done!
Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$\sum_{cyc}\frac{x}{x^2+(x+y+z)^2}\leq\frac{9}{10(x+y+z)}$$ or $$9\prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)\sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)\geq0,$$ which is $9w^6+A(u,v^2)w^3+B(u,v^2)\geq0$, which says that $$9\prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)\sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)\geq$$ $$\geq\frac{1}{81}\left(\sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)\right)^2$$ is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives $$(x-1)^2(x+2)^2(79x^2+253x+253)\geq0,$$ which is obviously true.
Done!
Threw come to mind.
The first is Lagrange Multipliers.
The second is to reason directly. Plus in $(x,y,z)$ and then plug in $(x-a,y+b,z+c)$ where $a+b+c=0$. You can use algebra to show that the former produces a larger total than the latter, and it follows that the maximum occurs when all the values are equal.
The third is that by symmetry. First observe that the max must have non-negative coordinates. That condition, plus $x+y+z\leq 1$ define a convex optimization problem, whose maximum must be at a corner of the convex hull. It's simple to check that of the corners, $x=y=z$ gives the largest value.