$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$

Question

$a, b, c ∈ \mathbb{R}+.$

WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1-y)+(1-z)=2 \to x+y+z=1.$

We want to prove $ax+by+cz \leq \sqrt{2}.$ This somewhat looks like Cauchy-Schwarz so I tried that: $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2.$ The problem becomes $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq 2,$ since $a,b,c,x,y,z>0.$

Expressing $a,b,c$ in terms of $x,y,z$: $(\frac {1}{x} + \frac {1}{y} + \frac {1}{z} - 3)(x^2+y^2+z^2)$

$= x+y+z+\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 2.$

$\to \frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 1.$ Stuck here. Thinking about using AM-GM but not sure how. Help would be greatly appreciated.

Answer 1

Cauchy-Schwartz ... \begin{eqnarray*} \left( \frac{a}{\sqrt{1+a^2}} \frac{1}{\sqrt{1+a^2}} + \frac{b}{\sqrt{1+b^2}} \frac{1}{\sqrt{1+b^2}} + \frac{c}{\sqrt{1+c^2}} \frac{1}{\sqrt{1+c^2}} \right)^2 \\\leq \left( \frac{a^2}{1+a^2} + \frac{b^2}{1+b^2} +\frac{c^2}{1+c^2} \right) \left( \frac{1}{1+a^2} + \frac{1}{1+b^2} +\frac{1}{1+c^2} \right)= 2. \end{eqnarray*}

Answer 2

$$\sqrt2-\sum_{cyc}\frac{a}{1+a^2}=\sum_{cyc}\left(\frac{\sqrt2}{3}-\frac{a}{1+a^2}\right)=\sum_{cyc}\left(\frac{\sqrt2}{3}-\frac{a}{1+a^2}+\frac{1}{2\sqrt2}\left(\frac{2}{3}-\frac{a^2}{1+a^2}\right)\right)=$$ $$=\sum_{cyc}\frac{(a-\sqrt2)^2}{2\sqrt2(1+a^2)}\geq0.$$

Answer 3

Let $$B:=\frac{1} {1+a^2} + \frac{1} {1+b^2} + \frac{1} {1+c^2}$$

From: $$A:=\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2$$

we get $A+B =3$ so $B =1$.

Now by Cauchy inequality we have $$A\cdot B \geq \big(\underbrace{\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2}}_{C}\big)^2$$

So we have $C^2\leq 2$ and we are done.

Answer 4

Continuing from where you stopped.

Multiply both sides by $xyz$ to clear out the denominators. Then multiple the first six terms by $x+y+z=1$, while the RHS with $(x+y+z)^2=1$ to get:

$$\sum_{\text{sym}}x^4y + \sum_{\text{sym}}x^3y^2 \ge 2\sum_{\text{cyc}}x^3yz + 2\sum_{\text{cyc}}x^2y^2z$$

This inequality is true, as we have $\sum_{\text{sym}}x^4y \ge 2\sum_{\text{cyc}}x^2y^2z$ and $\sum_{\text{sym}}x^3y^2 \ge 2\sum_{\text{cyc}}x^3yz$. These follow by summing the cyclic AM-GM inequalities:

$$x^4y + z^4y \ge 2x^2z^2y$$

$$\text{and}$$ $$x^3y^2 + x^3z^2 \ge 2x^3yz$$

Answer 5

Let $a,b,c>0$ $$\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}=2\rightarrow \dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\leq \sqrt{2}$$ $$\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}=\left( 1-\dfrac{1}{1+a}\right) +\left( 1-\dfrac{1}{1+b}\right) +\left( 1-\dfrac{1}{1+c}\right) =$$ $$=3-\left( \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)$$ $$\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\leq 2\leftrightarrow 3-\left( \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right) \leq \sqrt{2}\leftrightarrow$$ $$\leftrightarrow \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\geq 3-\sqrt{2}\leftrightarrow \dfrac{a}{a+a^2}+\dfrac{b}{b+b^2}+\dfrac{c}{c+c^2}\geq 3-\sqrt{2}\leftrightarrow$$ $$\leftrightarrow \dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\geq \dfrac{a}{a+a^2}+\dfrac{b}{b+b^2}+\dfrac{c}{c+c^2}\geq 3-\sqrt{2}\leftrightarrow$$ $$\leftrightarrow 2\geq 3-\sqrt{2}\quad \text{(true)}$$