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

Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$: $$\sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4}$$

So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm not sure if that's helpful at all.

(You might want to "rename" those to $x,y,z$ to make writing easier): $$\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4$$

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Do you know Lagrange Multipliers? – Will Jagy Feb 6 '12 at 23:25
Nope, I'm high school student, I only know basic inequalities (HM-GM-AM-QM, Cauchy-Schwarz-Bunyakowski, Jensen's, Minkowski, Schur and such). – Lazar Ljubenović Feb 6 '12 at 23:34
Well, maybe someone will answer by those methods. LM says that the extrema, and other critical points, of the functional occur when $x=y=z,$ which means $\pm(1/\sqrt 3, 1/\sqrt 3, 1/\sqrt 3).$ – Will Jagy Feb 6 '12 at 23:42
Suggestion: make substitution $x=r\cos\theta\sin\phi$, $y=r\sin\theta\sin\phi$, $z=r\cos\phi$. Let r = 1. Still not easy, but now a spherical trig problem. Perhaps trig identities would help. – daniel Feb 7 '12 at 9:47

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 $(x_1,x_2,x_3)=\left(\frac 1{\sqrt 3},\frac 1{\sqrt 3},\frac 1{\sqrt 3}\right)$.

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Very interesting catch about similarity with tangens formula. Thanks. – Lazar Ljubenović Feb 7 '12 at 11:07
@LazarLjubenović Just so you know, this substitution did not come out of nowhere. It is commonly used in calculus. See en.wikipedia.org/wiki/Tangent_half-angle_substitution – Potato Jul 19 '13 at 0:04

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.

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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.

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Here's a solution with tangent line method.

Note that $$\frac{x}{1+x^2} \le \frac{3\sqrt{3}}{16}(x^2+1) \iff \frac{1}{48}(3x-\sqrt{3})^2(\sqrt{3}x^2+2x+3\sqrt{3}) \ge 0 \text{ (true by discriminant) }$$

Now we have $$\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt{3}}{16}(x^2+y^2+z^2+3) = \frac{3\sqrt{3}}{4}$$ as desired. $\blacksquare$

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