Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $

Solution 1:

We assume that $x_i\geq 0$ and let $\theta_i\in \left(0,\frac{\pi}2\right)$ sucht that $x_i=\tan\frac{\theta_i}2$. We have $\sin(\theta_i)=\frac{2x_i}{1+x_i^2}$ and since $\sin$ in concave on $\left(0,\frac{\pi}2\right)$, we have $$\sum_{i=1}^3\frac{x_i}{1+x_i^2}=\frac 32\sum_{i=1}^3\frac 13\sin(\theta_i)\leq \frac 32\sin\frac{\theta_1+\theta_2+\theta_3}3.$$ We have, using the convextiy of $x\mapsto \tan^2 x$: $$\frac 13=\frac 13\sum_{i=1}^3\tan^2\frac{\theta_i}2\geq \tan^2\frac{\theta_1+\theta_2+\theta_3}6,$$ so $\tan\frac{\theta_1+\theta_2+\theta_3}6\leq \frac 1{\sqrt 3}$ and $\frac{\theta_1+\theta_2+\theta_3}6\leq \frac{\pi}6$. Finally $$\sum_{i=1}^3\frac{x_i}{1+x_i^2}\leq \frac 32\sin \frac{\pi}3=\frac{3\sqrt 3}4,$$ with equality if and only if $(x_1,x_2,x_3)=\left(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3}\right)$.

Solution 2:

The second derivative of the auxiliary function $$f(u)\ :=\ {\sqrt{u}\over 1+u}\qquad (u\geq0)$$ computes to $$f''(u)={3(u-1)^2 -4\over 4u^{3/2}(1+u)^3} \ <\ 0\qquad(0\leq u\leq 1)\ ;$$ whence $f$ is concave for $0\leq u\leq 1$. Putting $u_i:=x_i^2$ we therefore have $${1\over3}\sum_{i=1}^3{x_i\over 1+x_i^2}\leq\sum_{i=1}^3{1\over3}f(u_i)\ \leq\ f\Bigl({\sum_{i=1}^3 u_i\over3}\Bigr)=f\Bigl({1\over3}\Bigr)={\sqrt{3}\over4}\ ,$$ as claimed.

Solution 3:

Another solution. By the AM-GM inequality, we have $$ \frac{ 2 \cdot \dfrac{1}{\sqrt{3}} \cdot x_i } {1+x_i^2} \le \frac{\dfrac{1}{3} + x_i^2 }{1+x_i^2} = 1 - \frac{2/3}{1+x_i^2}, $$ But $-1/(1+z)$ is a concave function of $z = x^2$, so \begin{align} \frac{ \dfrac{2}{\sqrt{3}} x_i } {1+x_i^2} &\le \sum_{i=1}^3 \left[1 - \frac{2/3}{1+x_i^2}\right] \\ &\le 3 \left[1 - \frac{2/3}{1 + \dfrac{1}{3}\left(\sum_{i=1}^3 x_i^2\right) } \right] = \frac{3}{2}. \end{align} Multiplying both sides by $\sqrt3/2$, we get the desired result.